Thermal Energy Transfer Exam: Specific Heat & Carnot Cycle

Colonel Frank Seely School 3.6.2.1 Thermal Energy Transfer Q1.(a) Lead has a specific heat capacity of 130 J kg−1 K−1. Explain what is meant by this statement. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (1) (b) Lead of mass 0.75 kg is heated from 21 °C to its melting point and continues to be heated until it has all melted. Calculate how much energy is supplied to the lead. Give your answer to an appropriate number of significant figures. melting point of lead = 327.5 °C specific latent heat of fusion of lead = 23 000 J kg−1 energy supplied ................................................... J (3) (Total 4 marks) Q2.The Carnot cycle is the most efficient theoretical cycle of changes for a fixed mass of gas in a heat engine. The graph below shows the pressure−volume ( p−V) diagram for a gas undergoing a Carnot cycle of changes ABCDA. Page 1 Colonel Frank Seely School (a) (i) Show that during the change AB the gas undergoes an isothermal change. (3) (ii) Explain how the first law of thermodynamics applies to the gas in the change BC. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) Determine the ratio , Page 2 (3) Colonel Frank Seely School where TA is the temperature of the gas at A and TC is the temperature of the gas at C. ratio ..................................................... (3) (b) Show that the work done during the change AB is about 110 J. (2) (c) When running at a constant temperature, one practical engine goes through 2400 cycles every minute. In one complete cycle of this engine, 114 J of energy has to be removed by a coolant so that the engine runs at a constant temperature. The temperature of the coolant rises by 18 °C as it passes through the engine. Calculate the volume of the coolant that flows through the engine in one second. specific heat capacity of coolant = 3.8 × 10 3 J kg–1 K–1 density of coolant = 1.1 × 10 3 kg m–3 volume flowing in one second ................................................ m 3 (3) (Total 14 marks) Page 3 Colonel Frank Seely School Q3.A liquid flows continuously through a chamber that contains an electric heater. When the steady state is reached, the liquid leaving the chamber is at a higher temperature than the liquid entering the chamber. The difference in temperature is Δ t. Which of the following will increase Δt with no other change? A Increasing the volume flow rate of the liquid B Changing the liquid to one with a lower specific heat capacity C Using a heating element with a higher resistance D Changing the liquid to one that has a higher density (Total 1 mark) Q4.The temperature of a hot liquid in a container falls at a rate of 2 K per minute just before it begins to solidify. The temperature then remains steady for 20 minutes by which time all the liquid has all solidified. What is the quantity ? A B C 10 K–1 D 40 K–1 (Total 1 mark) Q5.(a) Define the specific latent heat of vaporisation of water. ........................................................................................................................ Page 4 Colonel Frank Seely School ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (b) An insulated copper can of mass 20 g contains 50 g of water both at a temperature of 84 °C. A block of copper of mass 47 g at a temperature of 990 °C is lowered into the water as shown in the figure below. As a result, the temperature of the can and its contents reaches 100 °C and some of the water turns to steam. specific heat capacity of copper = 390 J kgsup class="xsmall">–1 Ksup class="xsmall">–1 specific heat capacity of water = 4200 J kgsup class="xsmall">–1 Ksup class="xsmall">–1 specific latent heat of vaporisation of water = 2.3 × 10 6 J kgsup class="xsmall">–1 Before placement (i) After placement Calculate how much thermal energy is transferred from the copper block as it cools to 100 °C. Give your answer to an appropriate number of significant figures. thermal energy transferred ........................................... J Page 5 Colonel Frank Seely School (2) (ii) Calculate how much of this thermal energy is available to make steam. Assume no heat is lost to the surroundings. available thermal energy ........................................... J (2) (iii) Calculate the maximum mass of steam that may be produced. mass ......................................... kg (1) (Total 7 marks) Q6.A cola drink of mass 0.200 kg at a temperature of 3.0 °C is poured into a glass beaker. The beaker has a mass of 0.250 kg and is initially at a temperature of 30.0 °C. specific heat capacity of glass = 840 J kg–1K–1 specific heat capacity of cola = 4190 J kg–1K–1 (i) Show that the final temperature, Tf, of the cola drink is about 8 °C when it reaches thermal equilibrium with the beaker. Assume no heat is gained from or lost to the surroundings. (2) Page 6 Colonel Frank Seely School (ii) The cola drink and beaker are cooled from Tf to a temperature of 3.0 °C by adding ice at a temperature of 0 °C. Calculate the mass of ice added. Assume no heat is gained from or lost to the surroundings. specific heat capacity of water = 4190 J kg–1 K–1 specific latent heat of fusion of ice = 3.34 × 10 5 J kg–1 mass .......................................... kg (3) (Total 5 marks) Q7. An electrical immersion heater supplies 8.5 kJ of energy every second. Water flows through the heater at a rate of 0.12 kg s–1 as shown in the figure below. (a) Assuming all the energy is transferred to the water, calculate the rise in temperature of the water as it flows through the heater. specific heat capacity of water = 4200 J kg–1 K–1 answer = ....................................... K (2) Page 7 Colonel Frank Seely School (b) The water suddenly stops flowing at the instant when its average temperature is 26 °C. The mass of water trapped in the heater is 0.41 kg. Calculate the time taken for the water to reach 100 °C if the immersion heater continues supplying energy at the same rate. answer = ....................................... s (2) (Total 4 marks) Q8. In a nuclear reactor the mean energy produced by each uranium-235 nucleus that undergoes induced fission is 3.0 × 10 –11 J. In one pressurised water reactor, PWR, the fuel rods in the reactor contain 2.0 × 10 4 kg of uranium-235 and 40% of the energy produced per second is converted to 500 MW of electrical output power. It is assumed that all the energy produced in the reactor core is removed by pressurised water in the coolant system. The pressure of the water is approximately 150 times greater than normal atmospheric pressure. The water enters the reactor at a temperature of 275 °C ad leaves at a temperature of 315 °C. Under the operational conditions of the reactor the mean density of water in the coolant circuit is 730 kg m –3 and the specific heat capacity of water is approximately 5000 J kg–1 K–1. normal atmospheric pressure = 1.0 × 10 5 Pa molar mass of uranium-235 = 0.235 kg (a) The equation below gives one induced fission reaction that takes place in a reactor. (i) State the name of the particle represented by X. ............................................................................................................... (ii) State the proton and nucleon numbers represented by p and n. p .......................................................... Page 8 (1) Colonel Frank Seely School n .......................................................... (2) (b) (i) Calculate the number of fission reactions that occur in the reactor each second. number of fission reactions per second ............................................ (2) (ii) The reactor fuel rods contain 2.0 × 10 4 kg of uranium-235. Assume that all this uranium-235 could be used. Calculate the maximum time, in years, for which the reactor could operate. time .........................................................years (4) (iii) Suggest why it is not possible to use all the uranium-235 in the reactor fuel rods. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (c) Calculate the force exerted by the pressurised water on each square centimetre of Page 9 (2) Colonel Frank Seely School the wall of the reactor. force .........................................................N (2) (d) Calculate, in m3 s–1, the flow rate of the water through the PWR reactor. You will need to use data from the passage at the beginning of the question. flow rate ......................................................... m3 s–1 (4) (e) In a PWR the cooling water also acts as the moderator in the reactor and boron rods are used to control the power output. Describe the physical processes that take place in the moderator and control rods. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ Page 10 Colonel Frank Seely School (4) (Total 21 marks) Q9. An electrical heater is placed in an insulated container holding 100 g of ice at a temperature of –14 °C. The heater supplies energy at a rate of 98 joules per second. (a) After an interval of 30 s, all the ice has reached a temperature of 0 °C. Calculate the specific heat capacity of ice. answer = ............................J kg–1K–1 (2) (b) Show that the final temperature of the water formed when the heater is left on for a further 500 s is about 40 °C. specific heat capacity of water = 4200 J kg–1K–1 specific latent heat of fusion of water = 3.3 × 10 5 J kg–1 (3) (c) The whole procedure is repeated in an uninsulated container in a room at a temperature of 25 °C. State and explain whether the final temperature of the water formed would be higher or lower than that calculated in part (b). ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... (2) (Total 7 marks) Page 11 Colonel Frank Seely School Q10. Molten lead at its melting temperature of 327°C is poured into an iron mould where it solidifies. The temperature of the iron mould rises from 27°C to 84°C, at which the mould is in thermal equilibrium with the now solid lead. mass of lead = 1.20 kg specific latent heat of fusion of lead = 2.5 × 10 4 J kg–1 mass of iron mould = 3.00 kg specific heat capacity of iron = 440 J kg–1K–1 (a) Calculate the heat energy absorbed by the iron mould. answer = ..................................... J (b) (2) Calculate the heat energy given out by the lead while it is changing state. answer = ...................................... J (1) (c) Calculate the specific heat capacity of lead. answer = ...................................... J kg–1 K–1 (3) Page 12 Colonel Frank Seely School (d) State one reason why the answer to part (c) is only an approximation. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (1) (Total 7 marks) Q11. (a) Calculate the energy released when 1.5 kg of water at 18 °C cools to 0 °C and then freezes to form ice, also at 0 °C. specific heat capacity of water = 4200 J kg–1 K–1 specific latent heat of fusion of ice = 3.4 × 105 J kg–1 ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (b) (4) Explain why it is more effective to cool cans of drinks by placing them in a bucket full of melting ice rather than in a bucket of water at an initial temperature of 0 °C. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (Total 6 marks) Page 13 Colonel Frank Seely School Q12. In a geothermal power station, water is pumped through pipes into an underground region of hot rocks. The thermal energy of the rocks heats the water and turns it to steam at high pressure. The steam then drives a turbine at the surface to produce electricity. (a) Water at 21°C is pumped into the hot rocks and steam at 100°C is produced at a rate of 190 kg s–1. (i) Show that the energy per second transferred from the hot rocks to the power station in this process is at least 500 MW. specific heat capacity of water = 4200 J kg–1 K–1 specific latent heat of steam = 2.3 × 106 J kg–1 ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) The hot rocks are estimated to have a volume of 4.0 × 10 6 m3. Estimate the fall of temperature of these rocks in one day if thermal energy is removed from them at the rate calculated in part (i) without any thermal energy gain from deeper underground. specific heat capacity of the rocks = 850 J kg–1 K–1 density of the rocks = 3200 kg m–3 ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (7) (b) Geothermal energy originates as energy released in the radioactive decay of the uranium isotope 4.2 MeV. U deep inside the Earth. Each nucleus that decays releases Calculate the mass of U that would release energy at a rate of 500 MW. Page 14 Colonel Frank Seely School half-life of molar mass of U = 4.5 × 109 years U = 0.238 kg mol–1 ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (5) (Total 12 marks) Q13. In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper of mass 0.12 kg is heated in the flame for several minutes. The copper is then transferred quickly to a beaker, of negligible heat capacity, containing 0.45 kg of water, and the temperature rise of the water measured. specific heat capacity of water = specific heat capacity of copper = (a) 4200 J kg–1 K–1 390 J kg–1 K–1 If the temperature of the water rises from 15 °C to 35 °C, calculate the thermal energy gained by the water. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (b) (i) State the thermal energy lost by the copper, assuming no heat is lost during its transfer. Page 15 Colonel Frank Seely School ............................................................................................................. (ii) Calculate the fall in temperature of the copper. ............................................................................................................. ............................................................................................................. ............................................................................................................. (iii) Hence calculate the temperature reached by the copper while in the flame. ............................................................................................................. Q14. (4) (Total 6 marks) (a) Suggest two reasons why an α particle causes more ionisation than a β particle of the same initial kinetic energy. You may be awarded marks for the quality of written communication in your answer. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (b) A radioactive source has an activity of 3.2 × 10 9 Bq and emits α particles, each with kinetic energy of 5.2 Me V. The source is enclosed in a small aluminium container of mass 2.0 × 10–4 kg which absorbs the radiation completely. (i) Calculate the energy, in J, absorbed from the source each second by the aluminium container. Page 16 Colonel Frank Seely School ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) Estimate the temperature rise of the aluminium container in 1 minute, assuming no energy is lost from the aluminium. specific heat capacity of aluminium = 900 J kg–1 K–1 .............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (5) (Total 7 marks) Q15.A female runner of mass 60 kg generates thermal energy at a rate of 800 W. (a) Assuming that she loses no energy to the surroundings and that the average specific heat capacity of her body is 3900 J kg–1K–1, calculate (i) the thermal energy generated in one minute, ............................................................................................................. ............................................................................................................. (ii) the temperature rise of her body in one minute. ............................................................................................................. ............................................................................................................. ............................................................................................................. Page 17 Colonel Frank Seely School ............................................................................................................. (3) (b) In practice it is desirable for a runner to maintain a constant temperature. This may be achieved partly by the evaporation of sweat. The runner in part (a) loses energy at a rate of 500 Wby this process. Calculate the mass of sweat evaporated in one minute. specific latent heat of vaporisation of water = 2.3 × 10 6 J kg–1 ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3) (c) Explain why, when she stops running, her temperature is likely to fall. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (Total 8 marks) Q16. (a) A 2.0 kW heater is used to heat a room from 5 °C to 20 °C. The mass of air in the room is 30 kg. Under these conditions the specific heat capacity of air = 1000 J kg–1 K–1. Calculate (i) the gain in thermal energy of the air, ............................................................................................................. ............................................................................................................. (ii) the minimum time required to heat the room. Page 18 Colonel Frank Seely School ............................................................................................................. ............................................................................................................. ............................................................................................................. (4) (b) State and explain one reason why the actual time taken to heat the room is longer than the value calculated in part (a)(ii). ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (Total 6 marks) Q17.An electric shower heats the water flowing through it from 10°C to 42°C when the volume flow rate is 5.2 × 10−5 m3 s−1. (a) (i) Calculate the mass of water flowing through the shower each second. density of water = 1000 kg m−3 ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Calculate the power supplied to the shower, assuming all the electrical energy supplied to it is gained by the water as thermal energy. specific heat capacity of water = 4200 J kg−1 K−1. ............................................................................................................... ............................................................................................................... ............................................................................................................... Page 19 Colonel Frank Seely School ............................................................................................................... ............................................................................................................... (4) (b) A jet of water emerges horizontally at a speed of 2.5 m s −1 from a hole in the shower head. The hole is 2.0 m above the floor of the shower. Calculate the horizontal distance travelled by this jet. Assume air resistance is negligible. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3) (Total 7 marks) Q18. A bicycle and its rider have a total mass of 95 kg. The bicycle is travelling along a horizontal road at a constant speed of 8.0 m s –1. (a) Calculate the kinetic energy of the bicycle and rider. ...................................................................................................................... ...................................................................................................................... (2) (b) The brakes are applied until the bicycle and rider come to rest. During braking, 60% of the kinetic energy of the bicycle and rider is converted to thermal energy in the brake blocks. The brake blocks have a total mass of 0.12 kg and the material from which they are made has a specific heat capacity of 1200 J kg–1 K–1. (i) Calculate the maximum rise in temperature of the brake blocks. ............................................................................................................. ............................................................................................................. ............................................................................................................. Page 20 Colonel Frank Seely School ............................................................................................................. (ii) State an assumption you have made in part (b)(i). ............................................................................................................. ............................................................................................................. Q19.(a) (4) (Total 6 marks) A 3.0 kW electric kettle heats 2.4 kg of water from 16°C to 100°C in 320 seconds. (i) Calculate the electrical energy supplied to the kettle. ............................................................................................................... ............................................................................................................... (ii) Calculate the heat energy supplied to the water. specific heat capacity of water = 4200 J kg–1 K–1 ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) Give one reason why not all the electrical energy supplied to the kettle is transferred to the water. ............................................................................................................... ............................................................................................................... (4) (b) The potential difference supplied to the kettle in part (a) is 230 V. Page 21 Colonel Frank Seely School (i) Calculate the resistance of the heating element of the kettle. ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) The heating element consists of an insulated conductor of length 0.25 m and diameter 0.65 mm. Calculate the resistivity of the conductor. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (5) (Total 9 marks) Q20. A tray containing 0.20 kg of water at 20 °C is placed in a freezer. (a) The temperature of the water drops to 0 °C in 10 minutes. specific heat capacity of water = 4200 J kg–1 K–1 Calculate (i) the energy lost by the water as it cools to 0 °C, ............................................................................................................. ............................................................................................................. (ii) the average rate at which the water is losing energy, in J s –1. ............................................................................................................. ............................................................................................................. Page 22 (3) Colonel Frank Seely School (b) (i) Estimate the time taken for the water at 0 °C to turn completely into ice. specific latent heat of fusion of water = 3.3 × 10 5 J kg–1 ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) State any assumptions you make. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. Q21.(a) (3) (Total 6 marks) An electric shower heats water from 15°C to 47°C when water flows through it at a rate of 0.045 kg s–1. (i) Calculate the energy supplied to the water each second by the heating element in the shower. specific heat capacity of water = 4200 J kg–1 K–1 ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Show that the power of the heating element is 6.0 kW. Assume there is no heat loss to the surroundings. ............................................................................................................... Page 23 Colonel Frank Seely School ............................................................................................................... (3) (b) (i) The heating element in part (a) is connected to an alternating supply at 230 V rms. Calculate the rms current passing through the heating element in normal operation. ............................................................................................................... ............................................................................................................... (ii) The live wire and the neutral wire in the connecting cable are insulated copper wires of diameter 2.4 mm. Calculate the resistance per metre length of copper wire of this diameter. resistivity of copper = 1.7 × 10 −8 Ωm ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) Show that in normal operation, the potential drop per metre along the cable is 0.20 V m–1. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (iv) Electrical safety regulations require the potential drop along the cable to be less than 6.0 V. Calculate the maximum safe distance along the cable from the distribution board to the heating element. Page 24 Colonel Frank Seely School ............................................................................................................... ............................................................................................................... ............................................................................................................... (9) (Total 12 marks) Q22. (a) A student immerses a 2.0 kW electric heater in an insulated beaker of water. The heater is switched on and after 120 s the water reaches boiling point. The table below gives data collected during the experiment. initial mass of beaker initial mass of beaker and water initial temperature of water final temperature of water 25 g 750 g 20 °C 100 °C Calculate the specific heat capacity of water if the thermal capacity of the beaker is negligible. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (4) (b) The student in part (a) continues to heat the water so that it boils for 105 s. When the mass of the beaker and water is measured again, it is found that it has decreased by 94 g. (i) Calculate a value for the specific latent heat of vaporisation of water. ............................................................................................................. ............................................................................................................. (ii) State two assumptions made in your calculation. ............................................................................................................. ............................................................................................................. (4) (Total 8 marks) Page 25 Colonel Frank Seely School Q23.Use the following data to answer the question below. specific latent heat of fusion of lead = 23 kJ kg–1 molar mass of lead = 0.21 kg mol–1 (a) Estimate the mass of a lead atom. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (b) Estimate the energy supplied to an atom of lead when solid lead melts. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (c) Calculate the speed of a lead atom with the same kinetic energy as the energy supplied in part (b). ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (Total 5 marks) Q24.An electrical heater is used to heat a 1.0 kg block of metal, which is well lagged. The table shows how the temperature of the block increased with time. Page 26 Colonel Frank Seely School temp / °C 20.1 23.0 26.9 30.0 33.1 36.9 time / s 0 60 120 180 240 300 (a) Plot a graph of temperature against time on the grid provided. (3) Page 27 Colonel Frank Seely School (b) Determine the gradient of the graph. ........................................................................................................................ ........................................................................................................................ (2) (c) The heater provides thermal energy at the rate of 48 W. Use your value for the gradient of the graph to determine a value for the specific heat capacity of the metal in the block. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (d) The heater in part (c) is placed in some crushed ice that has been placed in a funnel as shown. The heater is switched on for 200 s and 32 g of ice are found to have melted during this time. Use this information to calculate a value for the specific latent heat of fusion for water, stating one assumption made. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3) (Total 10 marks) Page 28 Colonel Frank Seely School Q25.The following data refer to a dishwasher. power of heating element 2.5 kW time to heat water 360 s mass of water used 3.0 kg initial temperature of water 20°C final temperature of water 60°C (a) Taking the specific heat capacity of water to be 4200 J kg–1 K–1, calculate (i) the energy provided by the heating element, ............................................................................................................... ............................................................................................................... (ii) the energy required to heat the water. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (4) (b) Give two reasons why your answers in part (a) differ from each other. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ Page 29 (2) (Total 6 marks) Colonel Frank Seely School Q26.(a) State the principle of conservation of linear momentum for two colliding bodies. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (b) A bullet of mass 0.010 kg travelling at a speed of 200 m s–1 strikes a block of wood of mass 0.390 kg hanging at rest from a long string. The bullet enters the block and lodges in the block. Calculate (i) the linear momentum of the bullet before it strikes the block, ............................................................................................................... (ii) the speed with which the block first moves from rest after the bullet strikes it. ............................................................................................................... ............................................................................................................... (4) (c) During the collision of the bullet and block, kinetic energy is converted into internal energy which results in a temperature rise. Page 30 Colonel Frank Seely School (i) Show that the kinetic energy of the bullet before it strikes the block is 200 J. ............................................................................................................... (ii) Show that the kinetic energy of the combined block and bullet immediately after the bullet has lodged in the block is 5.0 J. ............................................................................................................... (iii) The material from which the bullet is made has a specific heat capacity of 250 J kg–1 K–1. Assuming that all the lost kinetic energy becomes internal energy in the bullet, calculate its temperature rise during the collision. ............................................................................................................... ............................................................................................................... ............................................................................................................... (5) (d) The bullet lodges at the centre of mass G of the block. Calculate the vertical height h through which the block rises after the collision. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (Total 13 marks) Q27.(a) Explain what is meant by (i) the specific heat capacity of water, ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... Page 31 Colonel Frank Seely School (ii) the specific latent heat of fusion of ice. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (b) A sample of solid material, which has a mass of 0.15 kg, is supplied with energy at a constant rate. The specific heat capacity of the material is 1200 J kg –1 K–1 when in the solid state. During heating, its temperature is recorded at various times and the following graph is plotted. Assume there is no heat exchange with the surroundings. (i) Show that energy is supplied to the material at a rate of 24 W. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Calculate the specific latent heat of fusion of the material. ............................................................................................................... Page 32 (4) Colonel Frank Seely School ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) Calculate the specific heat capacity of the material when in the liquid state. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (6) (Total 10 marks) Q28.(a) (i) State what is meant by thermal equilibrium. ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Explain thermal equilibrium by reference to the behaviour of the molecules when a sample of hot gas is mixed with a sample of cooler gas and thermal equilibrium is reached. ............................................................................................................... ............................................................................................................... ............................................................................................................... (3) (b) A sealed container holds a mixture of nitrogen molecules and helium molecules at a temperature of 290 K. The total pressure exerted by the gas on the container is 120 kPa. molar mass of helium molar gas constant R the Avogadro constant NA = = = Page 33 4.00 × 10–3 kg mol–1 8.31 J K–1 mol–1 6.02 × 1023 mol–1 Colonel Frank Seely School (i) Calculate the root mean square speed of the helium molecules. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Calculate the average kinetic energy of a nitrogen molecule. ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) If there are twice as many helium molecules as nitrogen molecules in the container, calculate the pressure exerted on the container by the helium molecules. ............................................................................................................... ............................................................................................................... ............................................................................................................... (6) (Total 9 marks) Q29. (a) (i) liquid. Explain what is meant by the specific latent heat of vaporisation of a ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) Suggest why the specific latent heat of vaporisation of water is much greater Page 34 Colonel Frank Seely School than the specific latent heat of fusion of water. ............................................................................................................... ............................................................................................................... ............................................................................................................... (3) (b) A cup contains 0.25 kg of water at a temperature of 15 °C. The water is heated by passing steam at 100 °C into it. specific heat capacity of water = 4200 J kg–1 K–1 specific latent heat of vaporisation of water = 2.3 × 106 J kg–1 boiling point of water = 100 °C (i) Use the above data to calculate the minimum mass of water that is in the cup when the temperature of the water reaches its boiling point. (ii) Explain why there is likely to be a greater mass of water in the cup than you have calculated in part (b)(i). ............................................................................................................... ............................................................................................................... ............................................................................................................... (4) (Total 7 marks) Q30. (a) A 500 µF capacitor and a 1000 µF capacitor are connected in series. Calculate the total capacitance of the combination. Page 35 Colonel Frank Seely School (2) (b) The figure below shows a diagram of an arrangement used to investigate the energy stored by a capacitor. The bundle of constantan wire has a resistance of 8.5 Ω. The capacitor is initially charged to a potential difference of 9.0 V by closing S1. (i) Calculate the charge stored by the 0.25 F capacitor. (ii) Calculate the energy stored by the capacitor. (iii) Switch S1 is now opened and S2 is closed so that the capacitor discharges through the constantan wire. Calculate the time taken for the potential difference across the capacitor to fall to 0.10 V. (7) Page 36 Colonel Frank Seely School (c) The volume of constantan wire in the bundle in the figure above is 2.2 × 10 –7 m3. density of constantan = 8900 kg m–3 specific heat capacity of constantan = 420 J kg–1 K–1 (i) Assume that all the energy stored by the capacitor is used to raise the temperature of the wire. Use your answer to part (b)(ii) to calculate the expected temperature rise when the capacitor is discharged through the constantan wire. (ii) Give two reasons why, in practice, the final temperature will be lower than that calculated in part (c)(i). ............................................................................................................... ............................................................................................................... ............................................................................................................... (5) (Total 14 marks) Q31. (a) Explain the meaning of the statement the specific heat capacity of ice is 2100 J kg –1K–1. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (b) An engineer is designing an ice-making machine. Water will enter the device at 18°C and the ice cubes are to be cooled to –5°C before release. (i) Show that about 0.4 MJ of energy must be removed from 1.0 kg of water at Page 37 Colonel Frank Seely School 18°C to change it into ice at –5°C. Set out the stages in your answer clearly. specific heat capacity of water specific heat capacity of ice specific latent heat of fusion of ice = 4.2 ×103 J kg–1 K–1 = 2.1 ×103 J kg–1 K–1 = 3.3 ×105 J kg–1 (3) (ii) The design brief requires that 1.5 kg of water is frozen in 300 s. Calculate the rate at which energy must be removed by the machine. (2) (Total 7 marks) Q32.In a power station, water is heated in a boiler to create steam. This steam is passed through a turbine before being cooled by river water to condense it back into the liquid state. The cooled water is then pumped back to the boiler for re-use. (a) The cooling water enters the condenser at 16°C and is returned to the river at 40°C. Every second, 35 × 103 kg of water are removed from the river. Calculate the power required to heat this water. Specific heat capacity of water, c, = 4200 J kg−1 K−1 (2) Page 38 Colonel Frank Seely School (b) Assume that all the energy transferred by the river water came from steam changing to water without a change in temperature. Calculate the mass of steam passing through the turbine per second. Specific latent heat of vaporisation of water = 2.4 × 10 6 J kg−1 (c) (2) The turbine is linked to a generator that produces 800 MW of electrical power. Calculate the efficiency of conversion of the internal energy of the steam into electrical energy. (2) (Total 6 marks) Q33.A meteorite of mass 1.2 × 104 kg enters the Earth's atmosphere at a speed of 2.5 km s –1. (a) Calculate the kinetic energy of the meteorite as it enters the atmosphere. (2) (b) The meteorite is initially at a temperature of 25°C. It is made of iron of specific heat capacity 110 J kg–1 K–1 and of melting point 1810°C. As the meteorite travels through the Earth’s atmosphere friction causes its temperature to rise. Calculate the energy needed to raise the temperature of the meteorite to its melting point. You should assume that the specific heat capacity of iron remains constant. (3) (c) When it reaches the surface of the Earth the mass of the meteorite has fallen to 5.5 × 103 kg and its speed to 150 m s–1 so that its kinetic energy is only 6.2 × 107 J. On striking the Earth this mass penetrates the Earth’s crust and is brought to rest in a distance of 25 m. Calculate: (i) the average force acting on the meteorite as it penetrates the Earth’s crust; (2) (ii) the time it takes for the meteorite to be brought to rest by the Earth’s crust; (2) (iii) the rate at which the meteorite dissipates energy in this time. (2) Page 39 Colonel Frank Seely School (Total 11 marks) Q34.The mass of a car and its passengers is 950 kg. When the brakes are applied the car decelerates uniformly from a speed of 25 m s –1 to a speed of 15 m s–1 in 2.5 s. (a) Calculate the decelerating force developed by the brakes. (2) (b) Calculate the work done in decelerating the car. (3) (c) Calculate the rate of energy dissipation by the brakes. (2) (d) There are four brake discs, each of mass 1.2 kg. The material from which the discs are made has a specific heat capacity of 510 J kg–1 K–1. (i) Assuming that all the energy dissipated during braking is converted into internal energy of the brake discs equally, calculate the temperature rise of the discs. (3) (ii) State and explain the effect on the temperature rise of one factor that has not been taken into account in the assumption in part (i). ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (2) (Total 12 marks) Q35.A car of mass M travelling at speed V comes to rest using its brakes. Energy is dissipated in the brake discs of total mass m and specific heat capacity c. The rise in temperature of the brake discs can be estimated from Page 40 Colonel Frank Seely School A B C D (Total 1 mark) Q36.A raindrop of mass m falls to the ground at its terminal speed v. The specific heat capacity of water is c and the acceleration of free fall is g. Given that 25% of the energy is retained in the raindrop when it strikes the ground, what is the rise in temperature of the raindrop? A B C D (Total 1 mark) Q37.Nitrogen at 20°C and a pressure of 1.1 × 10 5 Pa is held in a glass gas syringe as shown in Figure 1. The gas, of original volume 8.5 × 10 –5 m3, is compressed to a volume of 5.8 × 10–5 m3 by placing a mass on to the plunger of the syringe. The change in pressure of the gas is adiabatic. The new pressure of the gas is 1.9 × 10 5 Pa. Page 41 Colonel Frank Seely School Figure 1 (a) (i) Calculate the new temperature of the nitrogen. Give your answer in °C. (ii) Calculate the number of moles of nitrogen present in the syringe. (3) molar gas constant, R = 8.3 J mol–1 K–1 (3) (iii) The mass of the nitrogen in the syringe is 1.1 × 10 –4 kg. Calculate the mean square speed of the molecules when the gas has been compressed. (2) (iv) Explain why the change in pressure of the nitrogen is adiabatic. ............................................................................................................... ............................................................................................................... ............................................................................................................... (v) Explain, in terms of the behaviour of the nitrogen molecules, how the gas Page 42 (2) Colonel Frank Seely School exerts a greater pressure than it did before it was compressed. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (b) (3) After the adiabatic compression, the nitrogen is allowed to cool at constant volume. Figure 2 shows the variation of pressure with volume for the adiabatic compression and the subsequent cooling. The dotted line represents the isothermal compression that would have achieved the same final state. Figure 2 (i) Draw arrows on the graph to show the directions of the changes. Label your arrows adiabatic compression and cooling as appropriate. (1) (ii) State the significance of the shaded area of the graph. ............................................................................................................... ............................................................................................................... (1) (Total 15 marks) Page 43 Colonel Frank Seely School Q38.A 1.0 kΩ resistor is thermally insulated and a potential difference of 6.0 V is applied to it for 2.0 minutes. The thermal capacity of the resistor is 9.0 J K–1. The rise in temperature, in K, is A 1.3 × 10–3 B 8.0 × 10–3 C 0.48 D 0.80 (Total 1 mark) Q39.A thermometer has a thermal capacity of 1.3 J K–1. The initial temperature of the thermometer is 20°C. When used to measure the temperature of 40 g of water, it measures 37°C. (a) Determine the energy absorbed by the thermometer when it is placed in the water. (2) (b) Calculate the temperature change of the water as a result of introducing the thermometer. specific heat capacity of water = 4.2 × 10 3 J kg–1 K–1 Page 44 (2) (Total 4 marks) Colonel Frank Seely School M1.(a) (it takes) 130 J / this energy to raise (the temperature of) a mass of 1 kg (of lead) by 1 K / 1 °C (without changing its state) ✓ 1 kg can be replaced with unit mass. Marks for 130J or energy. +1 kg or unit mass. +1 K or 1 °C. Condone the use of 1 °K 1 (b) (using Q = mcΔT + ml) = 0.75 × 130 × (327.5 ‒ 21) + 0.75 × 23000 ✓ (= 29884 + 17250) = 47134 ✓ = 4.7 × 104 (J) ✓ For the first mark the two terms may appear separately i.e. they do not have to be added. Marks for substitution + answer + 2 sig figs (that can stand alone). 3 [4] M2.(a) (i) Appreciates pV should be constant for isothermal change (by working or statement) W =pΔV is TO Allow only products seen where are approximately 150 for 1 mark Penalise J as unit here M1 Demonstrates pV = constant using 2 points (on the line) set equal to each other or conclusion made or shows that for V doubling that p halves (worth 2 marks) need to see values for p and V Products should equal 150 to 2 sf Accept statement that products are slightly different so not quite isothermal A1 Demonstrates pV = constant using 3 points (on the line) with conclusion Need to see values for p and V Products should equal 150 to 2 sf Accept statement that products are slightly different so not quite isothermal Page 45 Colonel Frank Seely School A1 (ii) 3 Adiabatic therefore no heat transfer or Adiabatic therefore Q = 0 B1 Work is done by gas therefore W is negative or Work is done by gas therefore energy is removed from the system B1 ΔU is negative therefore internal energy of gas decreases or energy is removed from the system therefore internal energy of gas decreases or work done by the gas so internal energy decreases Allow −ΔU = −W or ΔU = −W B1 (iii) Uses pV / T = constant or uses pV=nRT or uses pV=NkT e.g. makes T subject or substitutes into an equation with pA and VAor pC and VC (condone use of n = 1) or their V read off range a = 2.5 to 2.6 (× 10−4) pA = 600 × 103 V read off range = 8.5 to 8.6 (× 10−4) pC = 140 × 103 C C1 Correct substitution of coordinates (inside range) into With consistent use of powers of 10 (pV) range is 150 to 156 and (pV) range is 119 to A C 120.4 C1 1.2(5) Allow range from 1.2 to 1.3 Page 46 3 Colonel Frank Seely School Accept decimal fraction : 1 A1 (b) 3 Energy per large square = 10(J) or states that work done is equal to area under curve (between A and B) or energy per small square = 0.4(J) or square counting seen on correct area Must be clear that area represents energy either by subject of formula or use of units on 10 or 0.4 Alternative: W = area of a trapezium (with working) or W = Pmean × ΔV or W = 450 × 103 × 2.5 × 10−4 or W = area of a rectangle + area of a triangle (with working) B1 Number of large squares = 10.5 to 11.5 seen and (W) = number of squares × area of one square (using numbers) Range = 105 to 115 (J) Or Number of small squares = 263 to 287 seen and (W) = number of squares × area of one square (using numbers) Range = 105 to 115 (J) States that actual work done would be lower because of curvature of line B1 (c) (Total energy removed per s =) 4560 (J) or number of cycles per s = 40 or (Mass per second =) 114 ÷ 68400 in rearranged form or their energy ÷ (c ΔT ) or their energy ÷ 68400 C1 0.067 (kg) seen Allow 0.066 (kg) here or allow V / t = 1.67 × 10−3 ÷ 1100 or ( )= and correct substitution seen Condone E = 114 (J) or temperature = 291(K) C1 = 0.061 × 10−3 or 6.06 × 10−5 (m3) Page 47 2 Colonel Frank Seely School A1 3 [14] M3.B [1] M4.A [1] M5.(a) the energy required to change the state of a unit mass of water to steam / gas ✓ when at its boiling point temperature / 100°C / without a change in temperature) ✓ allow 1 kg in place of unit allow liquid to vapour / gas without reference to water don't allow ‘evaporation’ in first mark (b) (i) (ii) thermal energy given by copper block ( = mcΔT) = 0.047 × 390 × (990 – 100) = 1.6 × 104 (J) ✓ 2 sig figs ✓ can gain full marks without showing working a negative answer is not given credit sig fig mark stands alone thermal energy gained by water and copper container ( = mcΔTwater + mcΔTcopper) = 0.050 × 4200 × (100 – 84) + 0.020 × 390 × (100 – 84) or = 3500 (J) ✓ (3485 J) available heat energy ( = 1.6 × 10 4 – 3500) = 1.3 × 104 (J) ✓ allow both 12000 J and 13000 J allow CE from (i) working must be shown for a CE take care in awarding full marks for the final answer – missing out the copper container may result in the correct answer but not be worth any marks because of a physics Page 48 2 2 Colonel Frank Seely School error (3485 is a mark in itself) ignore sign of final answer in CE (many CE’s should result in a negative answer) (iii) (using Q = ml) m = 1.3 × 104 / 2.3 × 106 = 0.0057 (kg) ✓ Allow 0.006 but not 0.0060 (kg) allow CE from (ii) answers between 0.0052 → 0.0057 kg resulting from use of 12000 and 13000 J 2 1 [7] M6.(i) (heat supplied by glass = heat gained by cola) (use of mg cg ΔTg =mc cc ΔTc) 1st mark for RHS or LHS of substituted equation 0.250 × 840 × (30.0 – Tf) = 0.200 × 4190 × (Tf – 3.0) 2nd mark for 8.4°C (210 × 30 – 210 tf = 838 Tf – 838 × 3) Tf = 8.4(1) Alternatives: 8°C is substituted into equation (on either side shown will get mark) resulting in 4620J~4190J or 8°C substituted into LHS (produces ΔT = 5.5°C and hence) = 8.5°C ~ 8°C 8°C substituted into RHS (produces ΔT = 20°C and hence) = 10°C ~ 8°C (ii) (heat gained by ice = heat lost by glass + heat lost by cola) NB correct answer does not necessarily get full marks (heat gained by ice = mcΔT + ml) heat gained by ice = m × 4190 × 3.0 + m × 3.34 × 105 (heat gained by ice = m × 346600) 3rd mark is only given if the previous 2 marks are awarded Page 49 2 Colonel Frank Seely School heat lost by glass + heat lost by cola = 0.250 × 840 × (8.41 – 3.0) + 0.200 × 4190 × (8.41 – 3.0) (= 5670 J) (especially look for m × 4190 × 3.0) the first two marks are given for the formation of the substituted equation not the calculated values m (=5670 / 346600) = 0.016 (kg) if 8oC is used the final answer is 0.015 kg or (using cola returning to its original temperature) (heat supplied by glass = heat gained by ice) (heat gained by glass = 0.250 × 840 × (30.0 – 3.0)) heat gained by glass = 5670 (J) (heat used by ice = mcΔT + ml) heat used by ice = m(4190 × 3.0 + 3.34 × 105) (= m(346600)) m (=5670 / 346600) = 0.016 (kg) 3 [5] M7. (a) 17 K 2 (b) t = 15 s 2 [4] M8. (a) (i) neutron B1 Page 50 1 Colonel Frank Seely School (ii) p = 36 B1 n = 144 B1 (b) (i) total energy produced = 2 MJ each second C1 number of reaction = 4.2 × 1019 per second A1 (ii) 2 1 kg contains (1000/235) × 6.02 × 10 23 atoms of uranium C1 total number of fissions = (1000/235) × 6.02 × 10 23 × 2 × 104 (5.1 × 1028) C1 time = total fissions available/number per second or 1.2 × 10 9s C1 38.7(39) years A1 (iii) 4 too few neutrons produced to maintain the chain reaction B1 probability of a neutron colliding with a uranium nucleus too low B1 more absorption of neutrons in non–fission capture B1 Page 51 2 Colonel Frank Seely School (c) pressure = 150 × 105 (Pa) or F = PA C1 force on 1 cm 2 = 1500N A1 (d) 2 energy removed each second E= MJ = 1.25 × 109 J or E = mc∆θ C1 1.25 × 109 = m 5000 × 40 C1 mass per second = 6250 kg C1 volume per second = 8.6(8.56) m 3 A1 (e) 4 control rods neutrons are absorbed B1 by the nucleus of the boron/atoms B1 moderator neutrons are slowed down B1 when colliding with the protons/hydrogen nucleus B1 4 [21] Page 52 Colonel Frank Seely School M9. (a) (use of ΔQ = m c ΔT) 30 × 98 = 0.100 × c × 14 c = 2100 (J kg–1 K–1) (b) 2 (use of ΔQ = m l + m c ΔT) 500 × 98 = 0.100 × 3.3 × 10 5 + 0.100 × 4200 × ΔT (ΔT = 38 °C) T = 38°C (c) 3 the temperature would be higher as the ice/water spends more time below 25°C or heat travels in the direction from hot to cold or ice/water first gains heat then loses heat any one line 2 [7] M10. (a) using Q = mcΔθ = 3.00 × 440 × (84-27) (1) 7.5 × 104 (J) (1) (b) 2 using Q = ml = 1.20 × 2.5 × 104 = 3.0 × 104 (J) (1) 1 Page 53 Colonel Frank Seely School (c) (heat supplied by lead changing state + heat supplied by cooling lead = heat gained by iron) 3.0 × 104 + heat supplied by cooling lead = 7.5 × 10 4 (1) heat supplied by cooling lead = 4.5 × 10 4 = mcΔθ c = 4.5 × 104/(1.2 × (327 – 84) (1) c = 154 (J kg–1 K–1) (1) (d) 3 any one idea (1) no allowance has been made for heat loss to the surroundings or the specific heats may not be a constant over the range of temperatures calculated 1 [7] M11. (a) (use of ΔQ = mcT gives) ΔQ1 = 1.5 × 4200 × 18 (1) = 1.134 × 105(J) (1) ΔQ2 = 1.5 × 3.4 × 105 = 5.1 × 105(J) (1) total energy released (= 1.134 × 10 5+ 5.1 × 105) = 6.2 × 105J (1) (6.23 × 105J) (b) (ice) requires energy to melt [or mention of latent heat] (1) stays at 0 °C (for longer) (or cools for longer) (1) (or extracts more energy from the drink) 4 2 [6] Page 54 Colonel Frank Seely School M12. (a) (i) heat water to 100 °C, energy (= 190 × 4200 × 79) = 63 (MJ) (1) vapourise water, energy (=190 × 2.3 × l06) = 440(MJ) (1) (437MJ) energy transferred (per sec) = (437 + 63) MJ (1) (= 500 MJ) (ii) mass of rocks (= 4.0 × 106 × 3200) = 1.3 × 1010(kg) (1) (1.28 × 1010) temperature fall of ΔT in one day, energy removed (= 1.28 ×1010 × 850 × ΔT) = 1.1 × 1013 ΔT (1) (1.09 x 1013 AT) (allow C.E. for value of mass of rocks) energy transfer in one day (= 500 × 10 6 × 3600 × 24) = 4.3 × 1013 (J) (1) in one day (b) K (1) number of nuclei in 1 kg of 238 U 7 = activity of lkg of 238U = (1) (1) (1) energy released per sec per kg of 238 U = 1.2(6) × 107 × 4.2 × 1.6 × 10–13(J) (1) (8.47 × 10–6(J)) mass of 238Uneeded = = 5.9(0) × 1013kg (1) 5 [12] Page 55 Colonel Frank Seely School M13. (a) (use of ∆Q = mc∆T gives) ∆Q = 0.45 × 4200 × (35 − 15) (1) = 3.8 × 104 J (3.78 × 104 J) (1) (b) 2 (i) 3.8 × 104 J (1) (allow C.E. for incorrect value of ∆Q from (a)) (ii) (mc∆T = ∆Q gives) 0.12 × 390 × ∆T = 3.8 × 104 (1) ∆T = 812 K (1) (use of ∆Q = 3.78 gives ∆T = 808 K (allow C.E. for incorrect value of ∆Q from (i)) (iii) (812 + 35) = 847 ºC (1) (use of 808 gives 843 ºC) (allow C.E. from (ii)) 4 [6] M14. (a) reasons: α particle has much more mass/momentum than β particle α particle has twice as much charge as a β particle α particle travels much slower than a β particle any two (1) (1) (b) (i) energy absorbed per sec (= energy released per sec) = 3.2 × 109 × 5.2 × 106 ×1.6 ×10–19 (1) = 2.7 ×10–3 (J) (1) (2.66 × 10–3 (J)) Page 56 2 QWC 1 Colonel Frank Seely School (ii) temperature rise in 1 minute = (for numerator) (1) (for denominator) (1) = 0.90 K (or °C) (1) (allow C.E. for incorrect value in (i)) 5 [7] M15.(a) (b) (c) (i) energy = 800 × 60 = 48 × 10 3J (1) (ii) (use of Q = mc θ gives) 48 × 103 = 60 × 3900 × θ (1) θ = 0.21 K (1) (0.205 K) (allow C.E. for value of energy from (i)) Q = ml gives 500 × 60 (1) = m × 2.3 × 106 (1) m = 0.013 kg (1) not generating as much heat internally (1) still losing heat (at the same rate) [or still sweating] (1) hence temperature will drop (1) 3 3 lmax 2 [8] M16. (a) (i) (use of ΔQ = mcΔθ gives) = 4.5 × 105 J (1) (ii) P × t = 4.5 × 105 (1) t= = 225 s (1) Page 57 Q = 30 × 1000 × 15 (1) Colonel Frank Seely School (allow C.E. for value of Q from (i) (b) 4 heat is lost to surroundings or other objects in room or to heater itself (1) more (thermal) energy required from heater (1) [or because convection currents cause uneven heating] [or rate of heat transfer decreases as temperature increases] 2 [6] M17.(a) (i) mass each sec [=(vol / sec) × density] = 5.2 × 10 –5 × 1000 (1) (1) (ii) = 0.052 kg (s–1) power (= energy supplied per sec = mcΔθ) = 0.052 × 4200 × (42 - 10) (1) = 7.0 × 103W (1) (6.99 × 103W) (allow C.E. for value of mass each sec from (i) (b) 4 h = ½gt gives the time to reach the floor (1) 2 = 0.64s (1) (0.639 s) range = (horizontal) speed of projection × time = 2.5 × 0.64 = 1.6 m (1) (allow C.E. for value of t) 3 [7] M18. (a) (use of Ek = ½mv2 gives) Ek = × 95 × 8.02 (1) = 3040 J (1) 2 Page 58 Colonel Frank Seely School (b) (i) ΔQ = 0.60 × 3040 = 1824 (J) (1) (allow C.E. for Ek from (a)) (use of ΔQ = mc Δθ gives) 1824 = 0.12 × 1200 Δθ (1) Δθ = 13 K (1) (12.7 K) (allow C.E. for ΔQ) (ii) no heat is lost to the surroundings (1) 4 [6] M19.(a) (i) (use of E = Pt gives) E = 3000 × 320 = 960 kJ (1) (ii) (use of Q = mcΔθ gives) Q = 2.4 × 4200 (100 - 16) (1) = 850 kJ (1) (iii) energy needed to heat the kettle material (1) [or heat loss to surroundings] 4 (b) (i) (use of I = I= gives) = 13 (A) (1) (use of V = IR gives) R = 18 Ω (1) (17.7 Ω) (allow C.E. for value of I) [or correct use of R = (ii) A=π (use of ρ = m) to give correct R] (m2) (1) (= 3.32 ×10–7(m2)) gives) (1) = 2.3 × 10-5 Ω m (1) (2.35 × 10-5 Ωm) (use of R = 18 Ω gives ρ = 2.4 × 10-5 Ω (allow C.E. for value of R from (i) and value of A) 5 [9] Page 59 Colonel Frank Seely School M20. (a) (i) (use of ΔQ = mcΔθ gives) energy lost by water = 0.20 × 4200 × 20 (1) = 1.7 × 104 J (1) (1.68 × 104 J) (ii) rate of loss of energy = = 28 (W) (1) (allow C.E. for value of energy lost in (i)) (b) 3 (i) (use of ΔQ = ml gives) (28 × t) = 0.20 × 3.3 × 105 (1) t = 2.4 × 103 s (1) (2.36 × 103 s) (allow C.E. for value of rate of loss of energy in (a)(ii) (ii) e.g. constant rate of heat loss (1) ice remains at 0°C (1) max 3 [6] M21.(a) (i) (in 1 s), E = 0.045 × 4200 × (47 – 15) (1) = 6050 J (ii) P = 6.0 kW (1) 3 (b) (i) (use of P = VI gives) I = = 26 A (1) (26.3 A) (allow C.E. for value of P from (a)) (ii) radius = 1.2 × 10-3 (m) (1) cross-sectional area = π(1.2 × 10-3)2 (or 4.5 × 10-6(m2)) (1) (1) = (1) Page 60 Colonel Frank Seely School = 3.8 × 10-3 Ω m-1 (allow C.E. for value of A) (iii) = 0.1 (V m–1) (per wire) two wires per cable gives pd per metre = 2 × 0.1 (1) (= 0.20 V m–1) (1) (iv) maximum length = = 30 m (1) 9 [12] M22. (a) (use of = Pt gives) 0.725 × c × (100 – 20) (1) = 2000 × 120 (1) c = 41 00 (1) J kg–1 (1) (4140 J kg–1) (b) (i) (use of mL = Pt gives) 94 × 10–3 L = 2000 × 105 (1) L = 2.2 × 106 J kg–1 (1) (ii) no evaporation (before water heated to boiling point) no heat lost (to the surroundings) heater 100% efficient any two (1) (1) 4 QWC 2 4 [8] M23.(a) mass of one atom = = 3.5 ×10–25 kg (1) (b) energy supplied = 23 × 10 3 ×3.5 ×10–25 (1) = 8.1 × 10–21 J (1) (8.05 × 10–21 J (allow C.E. for value from (i)) (c) (use of ½ mv2 = EK gives) ½ × 3.5 × 10–25 × v2 = 8.1 × 10–21 Page 61 Colonel Frank Seely School = 220 m s–1 (1) (215 m s–1) (EK = 8.05 × 10–21 gives v = 214 m s–1) (allow C.E. for value of EK from (ii)) M24.(a) (b) adequate scale (1) points plotted correctly (1) best fit line (at least a point to right and left of line) (1) 3 use of triangle for at least half line (1) gradient = 0.056 ± 0.004 (°C / s) (1) 2 gives 48 = c × (1.0) × 0.056 (1) (c) c = 860 ± 60 J kg K (or J kg °C ) (1) -1 (d) [5] -1 -1 -1 2 (use of Eth = ml gives) 48 × 200 = 32 ×10-3 × l (1) l = 3.0 × 105 J kg-1 (1) sensible assumption, e.g. no heat lost to surroundings or temperature does not change or heat is transferred to ice (1) 3 [10] M25.(a) (i) (P = VIt gives) P = 2500 × 360 (1) = 9.0 × 105 kJ (1) (ii) (Q = mcΔt gives) Q = 3 × 4200 × 40 (1) = 5 0 × 105 J (1) Page 62 (4) Colonel Frank Seely School (b) heat is lost to the surroundings the dishwasher is heated evaporation of the water any two (1) (1) (2) [6] M26.(a) (b) momentum before collision = momentum after collision (1) provided no external force acts (1) (i) p = mυ (1) 10 × 10–3 × 200 = 2.(0) (1) (ii) (c) (i) (2) kg m s–1 (N s) (1) total mass after collision = 0.40 kg (1) 0.40υ = 2.0 gives υ = 5.(0) m s–1 (1) (allow e.c.f. from (i)) (4) kinetic energy = ½mv2 (1) (= 200 J) (ii) kinetic energy = (iii) ΔQ = 200 – 5 = 195 (J) = mcΔθ (1) Δθ = (1) = 78 K (1) Page 63 (= 5.0 J) (allow e.c.f. for incorrect ΔQ) (5) Colonel Frank Seely School (d) kinetic energy lost (= potential energy gained) = mgh (1) 1.3 m (1) (2) [13] M27.(a) (i) quantity of energy supplied to unit mass (1) which raises temperature by 1°C [or 1K] (1) (ii) quantity of energy required to change state of unit mass (1) solid to liquid [or ice to water] (1) without change of temperature (1) (b) (i) Q (= mcΔθ ) = 0.15 × 1200 × (58 – 18) = 7200 (J) (1) P= (ii) (max 4) = 24 W (1) Q = 24 × 7 × 60 = 10080 (J) (1) 0.15l = 10080 gives l = 67200 J kg–1 (1) (iii) 24 × 4 × 60 = 0.15 × sL × (94 – 58) (1) gives sL = 1070 J kg–1 K–1 (1) (6) [10] M28.(a) (i) no net flow of (thermal) energy (between two or more bodies) (1) Page 64 Colonel Frank Seely School bodies at same temperature (1) (ii) (b) (kinetic) energy is exchanged in molecular collisions (1) until average kinetic energy of all molecules is the same (1) (i) max 3 (1) = 1340m s–1 (1) (ii) average k.e. of nitrogen molecules = average k.e. of helium molecules (1) = × (1340)2 = 5.97× 10–21 J (1) alternative schemes for (ii): average k.e. = = kT (1) × 1.38 × 10–23 ×290 = 6.00 × 10–21 J (1) or average k.e. = = (1) × = 6.00 × 10–21 J (1) (iii) use of p = molecules] (1) pHe = or equivalent [or, at same temperature, p ∝ no. of × 120 = 80 kPa (1) 6 [9] Page 65 Colonel Frank Seely School M29. (a) (i) energy/heat input needed to change liquid into gas/vapour when at its boiling point/without change of temperature M1 energy per unit mass/1 kg A1 (ii) idea that more energy has to be supplied to separate molecules than to break solid bond or for vaporisation work is done against atmospheric pressure or Idea that there is a greater change in PE in L-G than S-L B1 (b) (i) 3 ml = Mc∆θ or energy gain by water = 89250 (J) m × 2.3 × 106 = 0.25 × 4200 × 85 C1 m = 0.0388 kg A1 total mass = 0.289 (0.29) kg (0.25 + their m) B1 (ii) energy from steam is needed to raise temperature of the cup or energy/heat will be lost to the surroundings/cup/tube during the heating B1 4 [7] M30. (a) 1/C = 1/500 + 1/1000 or C1 Page 66 Colonel Frank Seely School 330 (333) µF A1 (b) (i) 2 Q = VC or Q = 0.25 × 9 C1 2.3 or 2.25 C (c.a.o. unit essential) A1 (ii) energy = ½ CV2 or 0.5 × 0.25 × 92 or ½ QV used C1 10(.1) J (allow e.c.f. for Q) A1 (iii) V = Vo e−t/RC C1 7 0.1 = 9 e−t/(8.5 x 0.25) C1 9.6 (9.56) s A1 (c) (i) Q = mc∆θ or mass = volume × density C1 correct substitution 10.1 = (2.2 × 10 −7 × 8900 × 400 × ∆θ) C1 12 (12.3) K or °C ecf for energy from (b) (ii) A1 Page 67 5 Colonel Frank Seely School (ii) some energy raises temperature of the thermometer B1 energy/heat lost to (raise temperature of) surroundings B1 [14] M31. (a) energy required to heat the ice up C1 2100 J needed to raise / extracted to lower temperature of 1 kg by 1 deg (K or °C) A1 (b) (i) 2 either water @ 18 to water @ 0 = 75600 J or ice @ 0 to ice at –5 = 10500 J M1 water @ 0 to ice @ 0 = 330000 J M1 total = 416100 J A1 (ii) 3 cand. bi × 1.5 or cand. bi/300 C1 power = 2080 W [0.4 MJ yields 2 kW condone 1 sf; J s–1] A1 2 [7] Page 68 Colonel Frank Seely School M32.(a) 35 × 103 × 4200 × 24 Cl = 3.53 × 109 W Al (b) 3.53 × 109 / 2.4 × 106 [ecf; ans to (a) / 2.4 × 106] = 1.47 × 103 kg s–1 [allow kg] (c) Cl Al 800 MW / sensible power Cl = 0.8 / (3.53 + 0.8) = 0.185 or 18.5 % [ecf from ai] Al [6] M33.(a) ½mv2 or substitution ignoring powers of 10 3.75 × 1010 J (b) A1 Q = mcΔθ C1 1785 seen C1 2.36 × 109 J (c) (i) C1 A1 W = Fs or correctly substituted values Page 69 C1 Colonel Frank Seely School 2.48 × 106 N condone effect of change of g.p.e. (ii) force = rate of change of momentum or use of an equation of motion A1 C1 0.33 s A1 (iii) P = W / t or P = Fv or substitution ignoring powers of 10 1.88 (or 1.86) × 108 W e.c.f. from (c)(ii) C1 A1 [11] M34.(a) Ft = Δ(mv) or F = ma and a = (v – u) / t C1 3800 N A1 (2) (b) work done = change in KE C1 or appropriate equation of motion for s or work done = Fs Calculation of one KE correctly or s calculated correctly (50 m) 1.9 × 105 J (condone N m) e.c.f. for F C1 A1 (3) (c) power = {force from (a)} × any velocity C1 or power = change in KE / time 76 kW (kJ s–1) ecf from (b) or ecf from (a) for use of P = Fv (their F × 20) A1 (2) Page 70 Colonel Frank Seely School (d) (i) (their (b)) = 4.8 (or 1.2) × 510 × Δθ (allow use of 1.2 instead of 4.8 for this mark) appreciation of 4 discs evident in the calculation 77.6 (78) K (or °C) or their (b) / 2450 C1 C1 A1 (3) (ii) temperature rise will be lower there will be air resistance some energy becomes internal energy of the air OR other components of the braking system (including answers involving friction of tyres with road) these will use some of the energy to increase temperature OR heat / energy transfer to the surroundings since surroundings at lower temperature or temperature or internal energy of surroundings rises B1 B1 B1 B1 (2) [12] M35.C [1] M36.D [1] M37.(a) (i) pV / T = constant in any form correct substitution including absolute temperatures / 345K 72°C not 345K Page 71 C1 C1 Colonel Frank Seely School condone no unit and condone just ° A1 (3) (ii) pV = nRT C1 correct substitution: n = or C1 3.8(5) × 10–3 (mol) or 3.8(4) × 10–3 (mol) A1 e.c.f. for their (i) (iii) pV = Nmc or p = 2 (3) ρc 2 3.0 × 105 m2 s–2 condone subsequent calculation of rms speed (iv) no heat transfer / ΔQ = 0 / no energy loss process too quick (for conduction to take place) / glass is poor (thermal) conductor / the system is isolated C1 A1 (2) B1 B1 (2) (v) molecules move faster / have more KE B1 greater number of collisions (per second) (between molecules and wall) not between molecules B1 greater (rate of) change in momentum in each collision B1 (3) (b) (i) anticlockwise arrows correctly labelled - both arrows needed B1 (3) Page 72 Colonel Frank Seely School (ii) work done on the gas (during compression) B1 (1) [15] M38.C [1] M39.(a) energy = heat capacity × temperature change 22 J C1 A1 (2) (b) E = mcθ 0.13 K condone °C / C1 (allow e.c.f. from (i)) A1 (2) [4] Page 73 Colonel Frank Seely School E1.This question was performed well by a majority of students. The explanation of a specific heat capacity in part (a) was very straightforward. The calculation in part (b) was done well by all but the weakest students even though it contained parts dealing with both specific heat capacity and latent heat. It was in choosing an incorrect number of significant figures that students lost the most marks. E2.(a) (i) There was a significant number of candidates who attempted this part by using W = PΔV or W = PV. They made statements that because the two work values were the same then there was no change in the internal energy. Other candidates thought that PV = temperature. Only the higher achieving candidates were able to produce convincing work that was correct and well-laid out with due regard to the fact that this was a “Show that...” style question. (ii) The application of the first law of thermodynamics to the adiabatic expansion was well understood and effectively communicated by most candidates. Some candidates were unaware that this expansion was adiabatic and neglected to address this in their explanation. (iii) Over 50% of candidates achieved all 3 marks in this part. However, in many cases the working was unconvincing with candidates opting for any supporting workings. A few candidates did not realise that have been without should expressed as a decimal fraction in keeping with standard procedure for numerical answers. (b) Many candidates attempted to use an approximation using W = Pmean × ΔV. This was equivalent to approximating the area under the curve to that of a trapezium. Such approaches were acceptable for 1 mark but required a statement identifying the answer as an over-estimate in order to access both marks. The usual method for estimating an area was not seen as often as expected and even when seen it was done in an unconvincing manner. Again, with a “Show that...” style question working had to be clearly communicated; many candidates were unable to do this effectively and consequently dropped marks. Page 74 Colonel Frank Seely School (c) Just under half of the candidates were able to deal with this multi-stage calculation and obtain all 3 marks. A common error was for candidates to convert the temperature increase of 18 °C to 291 K and substituting this value into Q = mcΔθ . Other common mistakes were an inability to deal with the per-second aspect of the question and also quoting or rearranging ρ = incorrectly. E5.A majority of candidates only scored one mark in part (a). These candidates either forgot to indicate a unit mass or, as in a majority of cases, they omitted the phrase, 'without a change in temperature', or equivalent. A few had problems in appreciating whether energy was required or whether energy was given out. It was very noticeable that at the lower ability end candidates have a poor vocabulary associated with this area of physics. Phrases like, 'to change water to a gas without changing state', or 'condense water into steam', and others showed a lack of distinction between boiling and evaporation. The calculation of part (b)(i) did not hold many difficulties for the bulk of the candidates but the significant figure issue did. In part (b)(ii) most candidates were relatively clear how to tackle this question. It was in the detail that errors were made. The most significant was to forget about the copper can, which also gained energy to reach the final temperature. Also at the lower ability end there were many opportunities to make arithmetic errors. The scores were much better for part (b)(iii) albeit from an error carried forward from part (b)(ii) in many cases. So the use of the latent heat equation is not difficult to grasp for a majority of candidates. The main error was from rounding off incorrectly or making errors in powers of 10 when converting to SI units. E6.Candidates found (i) quite difficult for a number of reasons. Some started correctly by equating heat supplied to glass equals heat gained by cola but then they could not make the final temperature the subject of the resulting equation. Others substituted the temperature the wrong way round and used (3 − T f), which was negative and fudged the arithmetic. As in a previous question candidates did not explain their approach which made it difficult to award partial marks. It was interesting to see some candidates who jumped in too quickly and made an initial mess of the calculation fared better on additional pages when they thought more carefully over the problem. Part (ii) was also very discriminating. Only the best candidates scored full marks. Good candidates who just missed full marks usually forgot about the 3 degree rise in temperature of the ice after it had melted. Most other candidates were aware of the mcΔT and ml equations but then made all manner of different errors. Page 75 Colonel Frank Seely School E7. E8. In part (a) almost all students knew the correct equation to use and only the less able students made errors. The first of these was to use the mass of water in the heating chamber rather than the rate of flow of water. The second error, which was less common, was to try to convert between Kelvin and Celsius by adding 273 to the answer. Again in part (b) it was only the less able students who had any difficulty. The problem was that they could not cope with being given the rate of supply of energy. Overall the question was done well. The neutron was identified by most students in part (a) (i). There were many correct answers to part (a) (ii) but less able students did not take account of the two ‘X‘ particles on the right hand side of the equation so obtained n= 145. Most made some progress with part (b) (i). Obtaining the total energy produced by the reactor proved difficult for less able students. There were a good proportion of correct answers in part (b) (ii). Some progress was made by a majority of the students. Few were able to make a sensible comment in part (b) (iii). Most were able to gain one of the two available marks in part (c) but the conversion from cm2 to m2 was the downfall of many students. Whilst many were able to gain credit in part (d) for making some progress, relatively few could correctly complete the problem. Most showed an appreciation of the different roles of the moderator in part (e) and the control rods but the majority gave inadequate detail of the processes. E9. Most candidates performed well in part (a). In part (b) the less able candidates tended to score only one mark because they could not form the energy balance equation when both changes of temperature and changes of state were taking place. Part (c) caught a majority of candidates out. Even grade A students were tempted to roll out the usual answer, ‘the temperature would be less because heat is lost to the surroundings’. This statement scored no marks. Page 76 Colonel Frank Seely School E10. Almost all candidates knew which equation to use in part (a) and only a small minority used the wrong temperature change. In part (b) most candidates obtained full marks. Part (c) turned out to be a very good discriminator. About one third of candidates were not using the heat energy released by the lead, as it cooled, in their calculation. These candidates either used their answer to (a) or (b) or the sum of the two. In addition another 10% calculated the incorrect temperature change. Part (d) was answered well on the whole. The most common error by candidates was to not say where the heat energy might go in their answer. Candidates simply said that heat is lost. E11. The calculation in part (a) was done well and candidates seemed to be quite happy calculating thermal energies. Some were still confused by the temperature scales and added 273 to the change in temperature, not realising that a difference in two temperatures on the Celsius scale will be in Kelvin. Part (b) generated some very interesting responses. It seems to be a common misapprehension that when ice melts it gives out energy and this cooled the cans of drinks. Candidates seem much happier discussing heat losses when something cools rather than when it warms up. Less able candidates discussed ice transferring ‘coldness’, and energy changes during a change of state is clearly a topic that is not well understood. E12. The calculation, in part (a) (i), of the energy needed to heat the water and to turn it to steam at 100°C was usually done correctly. Some failed by choosing an incorrect temperature change. In part (ii), many candidates obtained full marks although some correctly worked out the temperature change per second, but did not proceed to calculate the temperature change in one day. A small minority of candidates used an incorrect density formula or an incorrect value of the specific heat capacity. In part (b) most candidates knew how to convert MeV into joules correctly and used the given mass number correctly. They also knew how to calculate the total activity of the rocks from the energy released per second and the energy released per decay. Many of these candidates correctly used their activity value and the decay constant, worked out from the half-life, to calculate the total number of atoms and their mass. Some lost a mark in their calculation of the decay constant as a result of not converting the half-life into Page 77 Colonel Frank Seely School seconds. A significant number of candidates considered their answer for the total activity to be the total number of atoms and consequently lost two marks because they failed to use the total activity and the decay constant to calculate the total number of atoms. E13. Many candidates scored full marks in this question. Less able candidates created the usual confusion between the Kelvin and Celsius scale, but this was less of an issue than has been the case in the past. Candidates are clearly benefiting from practice with thermal energy questions. E14. Many answers in part (a) lacked sufficient detail to gain more than one or two marks. Although candidates knew that the mass of an α particle was greater than that of a β particle, few stated that its mass or momentum was much greater. Again, candidates knew that an α particle had more charge than a β particle but few stated it had twice as much charge. Few candidates realised that an α particle travels much slower than a β particle with the same kinetic energy; some even claimed that it travelled faster because it had more momentum. In addition, candidates often wrote at length about the relative penetrating powers of the two types of radiation. In part (b) (i), many candidates gave a correct calculation although some candidates made arithmetical errors, often in the conversion of MeV to J, or in rounding the answer incorrectly. In part (ii), the underlying principles behind the calculation were known but some candidates failed to realise that the increase in temperature in one minute was required. E15.Questions involving thermal energy have caused problems for candidates in previous papers, but this question was well answered. It also proved to be a good discriminator. Weaker candidates did have problems, however, in structuring their calculations in a logical way, leading to errors in the final answers. The explanation in part (c) was answered well by good candidates who, in some cases, provided considerable detail which was worthy of more than the two marks allocated. Page 78 Colonel Frank Seely School E16. This question was generally answered well with only a minority of candidates failing to complete the calculation. The explanation as to why actual heating might take longer generated a variety of explanations some of which were quite imaginative. E17.A large majority of the candidates gained full credit in part (a) (i) and were able to calculate the power correctly. However, the temperature difference confused other candidates, not realising that the temperature difference in °C is the same as in Kelvins. Some candidates lost the final mark as a result of a significant figure error or giving the power in joules and not watts or joules per second. Arithmetical errors leading to grossly unrealistic answers were made by a few candidates. In part (b), the time of descent needed to be calculated from the height drop in order to determine the horizontal distance travelled. Many candidates were able to do this correctly. Other candidates failed to gain any credit as a result of attempting to bring the speed of projection into the calculations for time of descent and the horizontal distance. E18. The last question in the paper was generally done well and candidates across the ability range were able to perform the various calculations successfully. A minority were confused by the need to use only 60% of the kinetic energy, but because marks were awarded for consequential errors, this did not prove to be too much of a penalty. E19.The majority of candidates were able to calculate the electrical energy supplied to the kettle in part (i) and the heat energy supplied to the water in part (ii). Significant figure penalties were imposed in part (ii). In part (iii), some candidates considered that the kettle cable became heated because of its resistance and many missed the point that there was heat loss from the kettle to the surroundings. In part (b) many candidates scored well, particularly in part (i). Candidates often lost a mark in part (ii) through an incorrect calculation of the cross-sectional area or by using an incorrect formula for the cross-sectional area. Weak candidates often penalised themselves heavily as a result of being unable to rearrange the resistivity equation correctly. Page 79 Colonel Frank Seely School E20. This question was well answered and candidates were consistently able to extract and use appropriate formulae in their calculations. In part (a), a minority of candidates used 293 K as the change in temperature rather than 20 K, but this was less prevalent than in previous years. There were problems with stating the appropriate assumptions in part (b). Many candidates stated incorrectly that no heat was lost to the surroundings. E21.Most candidates were awarded full marks in part (a) although some candidates added 273 to the temperature difference. A few candidates lost the mark in part (ii) because they failed to describe the correct relationship between power and energy. In part (b)(i), a significant number of candidates obtained the correct value of current but then divided, or multiplied, by , clearly unaware that the required answer had already been obtained. In part (ii), many candidates scored full marks with a clear and carefully expressed calculation, but a few candidates did lose one or more marks as a result of failure to use the correct value for the radius or the correct expression for the crosssectional area. A significant number of candidates were awarded only one mark in part (iii) because they had not obtained the correct value of resistance per metre in part (ii) or else they failed to appreciate that the cable contained two wires. In general, candidates who scored well in part (ii) usually scored both marks in part (iii). In part (iv), most candidates gained the available mark, including weaker candidates who were unable to make progress in the earlier parts of part (b). E22. A large number of candidates found this question difficult. Many seemed aware of the correct equation in part (a) but made mistakes such as using power instead of energy, converting temperatures to Kelvin and using an incorrect value for the mass of water. The latent heat calculation in part (b) produced even fewer correct responses. Many candidates included a change of temperature in their calculations or used an incorrect value for the mass of steam. Part b(ii) realised better answers although stating only one assumption was quite common. Page 80 Colonel Frank Seely School E23.Most candidates made progress in parts (a) and (b) and gained some credit. The most common error was due to the failure to recognise that specific latent heat was given in kJ kg –1 rather than J kg–1 . These candidates knew how to proceed in part (iii) but some used the molar mass instead of the mass of the atom or used the specific latent heat instead of the energy supplied to each atom. E24.Plotting the graph in part (a) was, by and large, well done and the majority of candidates were able to gain the allocated three marks. Again, calculating the gradient of the graph in part (b) was satisfactorily done although a minority of candidates found the inverse of the gradient. Parts (c) and (d) proved to be very good discriminators. The better candidates tackled the calculations impressively, but weaker candidates seemed unsure as to where to begin. The unit for specific heat capacity did cause a lot of problems to even the best candidates. E25.The calculations did not cause too many problems, but a minority of candidates were confused between the energy provided by the heating element and the energy required to heat the water and often the answers were interchanged. Part (b) proved to be more difficult and only a few candidates scored both marks available. E26.This question was very well answered with many candidates scoring maximum or nearly maximum marks. Most candidates knew in part (a) that the momentum before a collision equalled the momentum after the collision, but rather few gave the condition that no external force must act on the system. Part (b) was almost always correct, although some candidates did not add the masses of the bullet and block. Parts (c)(i) and (c)(ii) were usually correct. In section (iii) some candidates failed to subtract the remaining kinetic energy of 5.0 J from the initial kinetic energy of 200 J, or alternatively used 5.0 J for the internal energy. Part (d) was most often correct, but a number of candidates used ʋ2 = u2 + 2as and scored no marks. Page 81 Colonel Frank Seely School E27.It seemed that some centres had overlooked, or not had sufficient time to cover, the topic of heat calculations. Definitions of specific heat capacity and specific latent heat in part (a) were often very good and were awarded all four marks. A common misconception was that heat must be supplied to water in order to change it to ice. The calculations in part (b) were sometimes missed out completely or often misunderstood, but occasionally completed perfectly. Many candidates who were able to justify the 24W in part (b)(i) were then stumped by parts (b)(ii) and (b)(iii). In both of these parts the incorrect use of 7200J was very common. The answer to part (b)(iii) was often penalised because it was quoted to an excessive number of significant figures. E28.The anticipated emphasis in answers to part (a)(i) was on how thermal equilibrium is manifested, whilst in part (a)(ii) it was on a molecular explanation. In general, higher marks would have been awarded had the candidates read the question more carefully. Many answers simply fudged the two different aspects. Most candidates seem to have gained only the most basic understanding of molecular behaviour, a problem which pervaded the whole question. In particular, candidates should realise that the molecules in any system have a range of energies and that temperature is a measure of their mean kinetic energy. Thermal equilibrium may be satisfactorily explained by considering two systems in contact. When in equilibrium there is no net flow of energy between them because both systems are at the same temperature. However, there is a microscopic flow of energy in each direction between them. It is untrue that “every molecule has the same kinetic energy” when equilibrium has been established, as was stated in most answers. Very few candidates were able to calculate the r.m.s. speed of the helium molecules in part (b)(i). The main problems, also seen in other similar questions in the recent past, were confusion between molar and molecular quantities and an inability to differentiate between mean square values and r.m.s. values. Many candidates recovered in part (b)(ii) to gain a couple of marks by using 3 k T / 2 or its molar counterpart. Only a handful of candidates spotted that the average kinetic energy of a nitrogen molecule would be the same as that of a helium molecule, and that the result from part (b)(i) could be combined with ½ mʋ2 to arrive at the answer. There were many correct numerical answers to part (b)(iii) but most were arrived at instinctively; justification of (2 / 3) of 120KPa was expected for one of the two marks. E29. (a) (i) There were many good definitions, but many seemed unfamiliar with the idea of providing a formal definition/explanation. Many answers either Page 82 Colonel Frank Seely School forgot the ‘per kg’ or ‘at the boiling point’ aspect, so gained one of the two available marks. Poorer answers ignored the word specific completely, so that answers referred only to the energy to change from liquid to vapour. (b) E30. (ii) This was not done well and relatively few good answers were seen. Few seemed to have come across the idea of the energy needed to expand the vapour or gas against atmospheric pressure when boiling. Many wrote about it requiring more energy to break liquid bonds than to make solid bonds, apparently not appreciating that fusion is a melting process not a solidifying process. Answers that provided explanations in terms of more energy needed to break liquid bonds than to break solid bonds, without reference to the large increase in separation, were not convincing. (i) This was not done well and, surprisingly, few completely correct answers were seen. Many of those who found the mass of steam needed correctly then failed to add the 0.25 kg. Most knew the formulae to use but were at a loss to know what to do with the data. Often the 0.25 kg was used in the calculation of ml, and this was then equated to mcΔθ (giving an answer of 1.6 kg). Others simply used the 0.25 kg in ml and mcΔθ making no progress to the answer. (ii) Many candidates gave a satisfactory response. Those who did not often wrote about the condensed steam which had already been accounted for in (i). The vast majority of candidates found this a very accessible question and marks were generally very high. (a) The correct formula was used by most candidates. The main errors arising were failure to do the final reciprocal, inappropriate significant figures and units. (b) (i) This was usually successfully completed. (ii) This was also usually correct. (iii) Although there were many correct answers some confused V and Vo, others used the value of charge from (i) and some were unable to do the calculation having substituted correctly. Page 83 Colonel Frank Seely School (c) E31. (i) The majority knew which formula to use but a significant proportion did not know how to calculate the mass and some used the value for charge from (b) (i) for the energy (confusing Q in Q = VC and Q = mc∆θ). (ii) The majority gained credit for appreciating losses to the surroundings but few referred to the effect of the heat capacity of the thermometer which is the other most significant reason for the lower temperature. Some suggested that the lower temperature was because not all the heat energy supplied went to the thermometer. (a) Precise and clear explanations of the specific heat capacity statement were quite rare. There was often ambiguity in wording that was inappropriate at A2 level. (b) (i) The majority of candidates were able to carry out and structure this three-part calculation well to yield the correct answer. Examiners were lenient with those who presented their solution poorly, but candidates need to recognise that it is difficult to award full credit if information about the intermediate stages in the answer is missing. (ii) This was a simple calculation and was done well by all those who realised how straightforward it was. However, a minority felt the need to re-work only the latent heat energy part of the previous calculation and lost marks accordingly. E32.This question was answered well with many high scores of 5 or 6. (a) Some candidates carried out the evaluation using the temperature difference with 273 added to it. Significant figure and unit error penalties were very common. (b) Well done even by those who failed in (a). (c) A very common error was to forget to add the water power value into the total power involved in the system. Candidates who did this could still gain some credit Page 84 Colonel Frank Seely School however for a partially correct efficiency calculation. E33.(a) This part was generally well answered; the most common mistake was to quote the formula correctly but to forget to square the speed in the calculation. Weaker candidates forgot to convert km s−1 into ms−1. (b) Again this part was generally well answered. A minority of candidates added 273 onto the change of temperature before substituting their value into the specific heat capacity equation. (c) (i) A minority of candidates calculated the weight of the meteorite (using, F = mg). Candidates approaching this part using an equation of motion frequently calculated the maximum force rather than the average force. A considerable number of candidates used the value for time, calculated in part (ii) and substituted this into the relationship between force and rate of change of momentum. (ii) This part was generally well answered although many candidates suffered a significant figure penalty for quoting the time either as □ s or as (iii) s. A significant number of candidates made frequent use of P = Fv, but very few recognised that this meant that F needed to be the average force and v the average velocity in this example. E34.(a) This was usually completed successfully. A minority gave the deceleration rather than the decelerating force as the answer. (b) The majority decided to use the work done = Fs approach rather than the approach using Δ(½mv2) / t and there were a large number of correct answers to the question. Although the majority realised that they had to use work done = Fs, many failed to determine s correctly with a significant number arriving at a stopping distance of 75 m. These candidates used s = ut + ½at2 but used +4.0 m s–2 for a instead of –4.0 m s–2. (c) The majority appreciated that all that was necessary was to divide their answer to (b) by 2.5 s so that most were able to gain the two marks for this part allowing any error carried forward. (d) (i) This was often completed successfully. A common error was to use the value for rate of energy dissipation from part (c), instead of the total energy from part (b). The majority who were on the right track either divided the energy by 4 or multiplied the mass by 4 when using Q = mcΔθ. Calculating Δθ for one brake and then dividing by 4 was an unconvincing approach. Page 85 Colonel Frank Seely School (ii) E37.(a) (i) Many stated that the temperature rise of the brakes would be lower but were unable to give a convincing explanation. There were many who thought that friction when the brakes were applied had not been taken into account and that ‘energy would be lost to friction’. Others referred to energy dissipated due to friction between the tyres and the road but thought that this would increase the temperature rise of the brakes. Most candidates quoted the correct equation. The majority also carried on to complete the calculation correctly but a significant number used centigrade temperature instead of absolute temperature. (b) (ii) This was done correctly by most candidates. Some of those who had previously used centigrade temperatures corrected their mistakes here. (iii) Although the correct process was carried by most candidates, the use of incorrect data was common. Unit penalties and significant figure penalties were common on this part of the question. (iv) Many candidates explained what is meant by the word adiabatic but fewer could go on to say why the process described in the question would be approximately adiabatic. (v) Candidates tended to mention either the more frequent collisions with the wall of the syringe or the fact that the molecules moved faster. Relatively few mentioned both factors. Even fewer went further and explained how these factors would affect the rate of change of momentum of the molecules and, hence, the pressure exerted on the walls of the syringe. (i) This was generally well done. (ii) This was also answered correctly by many candidates but some were insufficiently clear about the direction of the process. Page 86 Colonel Frank Seely School E39.(a) Many candidates confused heat capacity with specific heat capacity. Otherwise, this calculation was well done. (b) Being more familiar to candidates, this calculation was done better than that in part (a). Those who calculated the energy incorrectly in (a) were not further penalised in this part. Page 87 Colonel Frank Seely School Resource currently unavailable. Page 88

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