1 THERMAL PHYSICS EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Heat, Temperature and Calorimetry Ans. (b) Temperature and Heat 1. At absolute zero (a) all substances exist in solid form (b) molecular motion ceases Sol. 3 KbT 0 2 t A 30 t 0 B 180 30 100 0 t A 30 t B 150 100 At critical temperature, the surface tension of a liquid is : (a) zero (b) infinity 4. (c) the same as that at any other temperature (d) cannot be determined Ans. (a) Sol. Surface tension decreases with the increases of temperature as temperature increases the molecular force decreases between the molecules. Surface tension vanishes at critical temperature. 3. t A H t A L t B H t B L So molecular motion ceases. 2. t B t B L t A H 180 C ; t B H 100C (b) As T = 0K i.e. absolute zero K.E. of molecules t A L 180 150 30C ; t B L 0C (c) water freezes (d) None of the above Ans. Sol. t A t A L The reading of Centigrade thermometer coincides with that of Fahrenheit thermometer in a liquid. The temperature of the liquid is : (a) –40°C (b) 0°C (c) 100°C Ans. Sol. C F 32 5 9 C=F=x x x 32 5 9 9x = 5x – 32 × 5 4x = –32 × 5 The graph between two temperature scales A and B is shown in figure. x 5. Between upper fixed point and lower fixed point, there are 150 equal divisions on scale A and 100 on scale B. The relationship for conversion between the two scales is given by : (a) t A 180 t B 100 150 (b) t A 30 t B 150 100 (c) t B 180 t A 150 100 (d) t B 40 t A 100 180 (d) 300°C (a) 32 5 40C 4 Which of the following parameters does not characterize the thermodynamic state of matter? (a) temperature (b) pressure (c) work (d) volume Ans. (c) Sol. The work does not characterize the thermodynamic state of matter. The radius of a ring is R and its coefficient of linear expansion is . If the temperature of ring increases by , then its circumference will increase by : (a) R (b) 2 R 6. (c) R Ans. (b) 2 (d) R 4 2 THERMAL PHYSICS Sol. L L0 1 T 2R1 2R 1 T = 2R +2R Hence cicumfrence increase by 2R A steel wire of cross–sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10–5/°C and its Young’s modulus is 2.0 × 1011 N/m2. (a) 24 N (b) 36 N (c) 12 N (d) 6 N Ans. (a) 7. Sol. 10. Equal masses of three liquids A, B and C have temperatures 10°C, 25°C and 40°C respectively. If A and B are mixed, the mixture has a temperature of 15°C. If B and C are mixed, the mixture has a temperature of 30°C. If A and C are mixed, the mixture will have a temperature of: (a) 16°C (b) 20°C (c) 25°C (d) 29°C Ans. (a) Sol. When A and B are mixed Tf = 15°C ms B 25 15 ms A 15 10 s A 2s B When B and C are mixed Tf = 30°C msB (30 – 25) = msC (40 – 30) 5sB = 10sC sB = 2sC ...(ii) From (i) & (ii) sA = 4sC When A and C are mixed Tf = ? F Y A L L F Y L A L ms A Tf 10 msC 40 Tf F Y T A s A Tf 10 sC 40 Tf F 2 1011 1.2 10 5 20 0.5 106 F = 24 N 8. A solid ball of metal has a spherical cavity inside it. If the ball is heated the volume of cavity will : (a) increase (b) decrease (c) remain unchanged (d) data insufficient Ans. (a) Sol. D D0 T Since T is positive therefore D ve Hence volume will increase. Thermal Expansion 9. An iron ball of mass 0.2 kg is heated to 10°C and put into a block of ice at 0°C. 2.5 g of ice melts. If the latent heat of fusion of ice is 80 cal/g, then the specific heat of iron in cal/g°C is : (a) 1 (b) 0.1 (c) 0.8 (d) 0.08 Ans. (b) Sol. m1s1T m 2 L 2 103 0.2 s iron 10 2.5 80 siron = 0.1 gm°C 4sC Tf 10 sC 40 Tf Tf = 16°C 11. The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed ? (a) 15.6 °C (b) 23.2 °C (c) 20.3 °C (d) 25.8 °C Ans. (c) Sol. When A and B are mixed msA (16 – 12) = msB (19 – 16) 4sA = 3sB ...(i) When B and C are mixed msB (23 – 19) = msC (28 – 23) 4sB = 5sC ...(ii) From equation (i) and (ii) 5 4s A 3 s C 4 When A and C are mixed ms A Tf 12 msC 28 Tf 3 THERMAL PHYSICS 3 5 s C Tf 12 s C 28 Tf 4 4 Then for same temperature 15Tf 180 28 16Tf lB l0 B T lA l0 A T Tf = 20.3°C 12. 50 g of ice at 0°C is mixed with 50 g of water at 60°C, final temperature of mixture will be : (a) 0°C (b) 40°C (c) 10°C (d) 15°C Ans. (a) Sol. When 50 gm water at 60°C is converted into 0°C water then heat released Qreleased = 50 × sw × (60 – 0) Qreleased = 3000 cal Suppose this heat can melt x gm of ice 3000 cal = xL x 3000 gm 37.5 gm 80 Thus this heat is not sufficient to melt 50 gm of ice. Hence final temperature will be ‘O°C’. 13. Two metal rods A and B are having their initial length in the ratio 2 : 3 and the co-efficients of linear expansion in the ratio 3 : 4. When they are heated through the same temperature difference, the ratio of their linear expansion is (a) 3 : 4 (b) 1 : 2 (c) 2 : 3 (d) 4 : 3 Ans. (b) Sol. l A l B lA 2 A 3 ; lB 3 B 4 As lB lA ; so will bend with B on outer side. 15. Ans. Sol. 1 2 Y1 lA lA A T lA A 2 3 1: 2 lB lB B T lB B 3 4 14. Ans. A bi-metallic strip is made of two strips A and B having co-efficients of linear expansion A and B. If A <B, then on heating the strip will (a) bend with A on outer side (b) bend with B on outer side (c) not bend at all (d) None of these (b) l1 l Y2 2 l1 l2 Y1 1T Y2 2 T Y1 2 3 3: 2 Y2 1 2 16. Ans. Sol. Two rods of different materials having coefficients of thermal expansion 1 and 2 and Young’s moduli Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that these undergo the same increase in temperature. There is no bending of the rods. If 1 : 2 = 2 : 3, the thermal stresses developed in the two rods are equal, provided Y1 : Y2 is (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9 (c) Given stress are equal Which of the following qualities are best suited for a cooking utensil (a) high specific heat and low thermal conductivity (b) high specific heat and high thermal conductivity (c) low specific heat and low thermal conductivity (d) low specific heat and high thermal conductivity (d) 4 THERMAL PHYSICS Sol. Heat flow: Q T KA t x Heat flow: Q K t Sol. Ans. Sol. 4 3 R 3 V R 3 V R V 3T V Heat absorbed : Q msT 17. V For cooking heat flow must be higher than heat absorbed so K must be high and s must be low for cooking. What should be the lengths of steel and copper rods at 0°C so that the length of the steel rod is 5 cm longer than the copper rod at any temperature? (Steel) = 1.1 × 10–5 °C–1 and (Copper) = 1.7 × 10–5 °C–1 (a) 14.17 cm, 9.17 cm (b) 9.17 cm, 14.17 cm (c) 28.34 cm, 18.34 cm (d) 14.17 cm, 18.34 cm (a) Let length of copper and steel rods be lcu and lsteel lsteel – lcu = 5 ...(i) At any temperature T, lcu lcu cu T lcu cu T 0 lsteel lsteel steel T lsteel steel T 0 4 V 3T R 3 3 V 4R 3T Numerical Value Type Questions 19. Steel wire of length L at 40°C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length L. The coefficient of linear thermal expansion of steel is 10–5/°C. Young’s modulus of steel is 1011N/m2 and radius of the wire is 1 mm. Assume that length (L) >> diameter 9d) of the wire. Then the value of ‘m’ (in kg to nearest, integer). Ans. 3.00 Sol. lcu lcu cu T F L mg Y Y A L A m AY g lsteel lsteel steel T For constant difference in length, change in length must be same at all temperature. lsteel lcu 20. 18. Ans. g 10 –3 1011 10–5 10 10 3kg Two rods of different materials having coefficients of thermal expansion 1 , 2 and Young’s moduli Y1, Y2 lsteel 1.1 10 lcu 1.7 10 11 lsteel 17 lcu r 2 Y 2 lsteel steel T lcu cu T 5 respectively are fixed separately between two rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. 5 ...(ii) From equation (i) and (ii) lsteel = 14.17 cm lcu = 9.17 cm The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal is . The sphere is heated a little by a temperature T so that its new temperature is (T + T). The increase in the volume of the sphere is approximately : (a) 2R T (b) R2 T (c) 4R3 T/3 (d) 4R3 T (d) If 1 : 2 = 2 : 3; then thermal stress developed in both Y1 rods would be equal, if Y is 2 Ans. 1.50 Sol. Thermal stress is given by Y Here is same and 1 : 2 2 : 3 So, Y11 Y2 2 for the stress to remain the same then we Y1 2 3 have Y 2 2 1 5 THERMAL PHYSICS 21. As a result temperature rise of 32°C, a bar with a crack at its center buckles upwards. If the fixed distance L0 is 4 m, and the coefficient of linear expansion of bar is 25 × 10–6 °C–1. Find the rise x (in cm, to the nearest integer) of the center. Sol. As the track won’t be allowed to expand linearly, the rise in temperature would lead to developing thermal stress in track. Stress y Y or YT Energy stored per unit volume 1 2 2 Y Energy stored per unit length A2 2Y A Y 2 T 2 2 10 –2 1011 10 –10 100 5J m 2 Calorimetry 24. Relation between molar and principal specific heat of gases (a) Cp = Mcp (b) Cp = M + cp (c) cp = MCp (d) Cp = M – cp Ans. (a) Sol. dQ = mcpdT ...(i) where cP = principal specific heat of substance m = mass of substance dQ = nCpdT ...(ii) Where Cp = molar specific heat at constant pressure n = number of moles from equation (i) and (ii) mcpdT = nCpdT Ans. 8.00 Sol. 2L L0 1 T 4 1.0008 L 2.0016 By Pythagoras theorem, x2 2.0016 22 x 2 2 1 8 22. x2 0.0008 x 0.8m 8 Two vessels connected at the bottom by a thin pipe with a sliding plug contain liquid at 20°C and 80°C respectively. The coefficient of cubic expansion of liquid is 10–3 K–1. h 20 The ratio of heights columns in the vessel h is nearest 80 Ans. Sol. to which integer? 1.00 23. Ans. 25. Because the sliding plug stay in the connecting pipe, the pressure in both the vessels at he level of hte pipe must be the same. h 20 d 20 h 80 d 80 d 0 1 – 80 d 0 1 – 20 h 20 d80 h 80 d 20 0.94 The area of cross - section of a railway track is 0.01 m2. The temperature variation is 10°C. Coefficient of linear expansion of material of track is 10–5/°C. The energy stored per meter in the track is _____ J/m. (Young’s modulus of material of track is 1011 Nm–2) 5.00 m Cp M Cp = Mcp Liquids at temperature 60°C and 20°C, respectively, have mass ratio 3 : 4 and their specific heats in the ratio 4 : 5. If the two liquid mixed, the resultant temperature (a) 70°C (b) 0°C (c) 35°C (d) 40°C (c) Let us consider final temperature is T. Heat given = Heat taken mc p Ans. Sol. m1s1T1 m 2 s 2 T2 m1 s1 60 T T 20 m2 s2 3 4 60 T T 20 4 5 180 – 3T = 5T – 100 8T = 280°C T = 35°C 6 THERMAL PHYSICS 26. Ans. Sol. 27. Heat given to a body which raises its temperature by 1°C is (a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient (b) Thermal capacity If m is the mass, is temp. and ‘s’ is specific heat, then thermal capacity K is given by (a) K = ms (b) K = m (c) K Ans. (d) Sol. K ms ...(1) Q ms T ...(2) Sol. Q mL L Q Since lenght of horizontal line (time) m for A is greater than B and heat is supplied at the same rate therefore QA < QB Q mL L Q msdT S Q for same time mdT dTA dTB SA SB Q T Q K T (d) LA LB 30. Ans. 29. Sol. (d) K = ms From (1) & (2) K = ms Dimensions of latent heat are (a) [M1 L2 T–2] (b) [M0 L2 T–2] (c) [M1 L1 T–1] (d) [M1 L 1 T–2] (b) 28. Ans. Ans. Sol. 31. Q ML2 T 2 M0 L2 T 2 M m Equal masses of two liquids A and B contained in vessels of negligible heat capacity are supplied heat at the same rate. The (temperature vs time) graphs for the two liquids are shown in figure. If S represents specific heat and L represents latent heat of liquid, then If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should (a) increase (b) remain unchanged (c) decrease (d) first increase then decrease (c) According to mass energy equivalence, when water is cooled to form ice, it lose some energy, hence its mass will decrease accordingly. Three copper blocks of masses M 1, M 2 and M 3 kg, respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at temperatures T1, T2, T3 (T1 > T2 > T3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is : (s is specific heat of copper) (a) T T1 T2 T3 3 (b) T M1T1 M 2 T2 M 3 T3 M1 M 2 M 3 (c) T M1T1 M 2 T2 M 3 T3 3 M1 M 2 M 3 (d) T M1T1s M 2 T2 s M 3 T3s M1 M 2 M 3 Ans. (b) Sol. Q system = 0 Q1 Q 2 Q3 0 M1S T T1 M 2S T T2 M3S T T3 0 M1T M1T1 M 2 T M 2 T2 M3 T M 3 T3 0 (a) SA > SB ; LA < LB (c) SA < SB ; LA < LB (b) SA > SB ; LA > LB (d) SA < SB ; LA > LB T M1T1 M 2 T2 M 3T3 M1 M 2 M 3 7 THERMAL PHYSICS Numerical Value Type Questions 32. In two experiments with a continuous flow - calorimeter to determine the specific heat capacity of a liquid, an input power of 16 W produced a rise of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of liquid three times faster. Find the power lost (in W) to the surroundings in each case, assuming that heat lost to surroundings depends on temp of liquid and atmospheric temp. only. Ans. 8.00 Sol. 35. cal A 50 gm lead bullet (specific heat 0.020 gC ) is initially Ans. Sol. at 30°. It is fired vertically upwards with a speed of 84 m/ s. On returning to the starting level, it strikes a slab of ice kept at 0°C. (A × 100) mg of ice is melted due to this. Find the value of ‘A’. (Take:- Lice = 80 cal/gm and 1 cal = 4.2 J) 9.00 Heat lost = Heat gain Let power lost to surrounding be Q. 1 mv 2 ms mice Lice 2 dm 16 – Q S 10 dt 1 50 10 –3 84 84 4.2 50 0.02 30 2 4.2 mice 80 dn and 32 – Q 3 s S 10 dt On simplification m ice 0.9 gm 900 mg 32 – Q 3 Q 8W 16 – Q 33. Ans. Sol. 34. A9 A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75° C. (given room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C. Temperature T (in °C) is: 885.00 100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45 100t – 750 = 450+ 7650 10t = 8850 t = 885°C Two liquids A and B are at 32° C and 24°C respectively. When mixed in equal masses the temperature of the mixture 36. Ans. Sol. SA if found to be 28°C. The ratio of specific heats S is B Ans. 1.00 Sol. Temperature of mixture mix L 270 S 37. A SA BSB SA SB Ans. Sol. 32 SA 24 SB 28 SA Ss 28SA 28SB 32SA 24SB SA 1 SB 1 A liquid at 30°C is poured very slowly into an open Calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporated. After pouring another 80 gm of the liquid the equilibrium temperature is found to be 50 °C. The ratio of the Latent heat of the liquid to its specific heat will be ____ °C. (Neglect the heat exchange with surrounding) 270.00 Calorimeter is open and after evaporation liquid escapes 5 × S × (80 – 30) + 5L = W × (110 – 80) 80 × S × (50 – 30) = W × (80 – 50) 5 × S × 50 + 5L = 80 × S × 20 5L= 1350 S How much heat is required to convert 8.0 g of ice at – 15°C to steam at 100°C? (Given, Cice = 0.53 cal/g–°C, Lf = 80 cal/g and Lv = 539 cal/g, and cwater = 1 cal/g–°C) Give the answer in kcal correct to one decimal place. 5.8 Q1 mcice Tf – Ti 8 THERMAL PHYSICS 8.0 0.53 0 – –15 63.6 cal 39. Q 2 mL f 8 80 640cal Q3 mcwater Tf – t i 8.0 0.1 100 – 0 800cal Q 4 mL v 8.0 539 4312 cal Net heat required, Q Q1 Q 2 Q3 Q 4 Ans. Sol. 5815.6cal 38. Ans. Sol. Ti 227 273 500 K The temperature of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. the temperature when A and B are mixed is 16° C and when B and C are mixed it is 23°C. What should be the temperature when A and C are mixed (in °C, correct to two decimal places)? 20.26 Let m be the mass of each liquid and SA, SB, SC specific heats of liquids A, B and C respectively. When A and B are mixed. The final temperature is 16°C. Heat gained by A = Heat lost by B And final temperature of container, Tf 27 273 300 K Now, heat gained by the ice cuibe = heat lost by the container 0.1 8 10 4 0.1 103 27 –m i.e., ms B 23 –19 msC 28 – 23 4 i.e.,SC SB 5 300 Ans. After substituting the values of A and B and the proper limits, we get m=0.495 kg m grams of steam at 100°C is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of m (in grams) is _______. 40.00 Sol. Here, heat absorbed by ice = mice Lf mice Cw 40 – 0 40. ...(2) 4 4 16 From Eqs. (i) and (ii), SC SA SA 5 3 15 A BT dT BT 2 –m AT or 10700 2 500 ...(1) When B and C are mixed. Heat gained by B = Heat lost by C 300 500 i.e.ms A 16 –12 ms B 19 –16 4 i.e.s B SA 3 An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A = 100 cal/kg-K and B = 2 × 10–2 cal/kg-K2. If the final temperature of the container is 27°C, determine the mass of the container (in kg, correct to one decimal place). (Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal/ kg-K). 0.5 Let m be the mass of the container. Initial temperature of container, When A and C are mixed, let the final temperature be Heat released by steam msteam L v msteam Cw 100 – 40 Heat gained by A = Heat lost by C Heat absorbed = heat released ms a –12 msC 28 – mice Lf mice Cw 40 – 0 msteam L v msteam C w 100 – 40 i.e. – 12 16 28 – 15 By solving, we get 628 20.26C 31 200 80 cal g 200 1cal g C 40 – 0 m 540cal g 540 1cal g C 100 – 40 200 80 40 1 m 540 60 1 m 40 g 9 THERMAL PHYSICS 41. 1 kg ice at –10°C is mixed with 0.2 kg of steam at 200°C. If final temperature (°C) of mixture at equilibrium is K1 T K 2 100 T 255 Teq , then fill the value of x. x Ans. Sol. Latent heat of fusion of ice = 80 cal/gram, latent heat of vaporization of water = 540 cal/gram, specific heat of water = 1 cal/gm K specific heat capacity of ice specific heat of steam = 0.5 cal/gram-K. 6.00 44. 1 1 10 1 80 1 Teq 2 1 0.2 100 0.2 540 0.2 1100 – Teq 2 Teq Ans. Sol. 265 6 Heat Transfer Heat Transfer, Conduction and Convection 42. Dimension of co-efficient of thermal conductivity are (a) [L0M1T–3K–1] (b) [L1M1T–3K–1] 1 1 –3 (c) [L M T K] (d) [L1M–1T–2K–1] Ans. (b) Sol. 2 T 3 100 T 45. Q dT KA t dx K1 K 2 (a) K K ML2 T 2 L K 2 L K T 1 Ans. Sol. Two metal rods A and B of equal lengths and equal cross sectional areas are joined end-to-end. The co-efficients of thermal conductivity of A and B are in the ratio 2 : 3. When the free end of A is maintained at 100°C and the free end of B is maintained at 0°C, the temperature of the junction is (a) 30°C (b) 40°C (c) 50°C (d) 60°C (b) Let temperature of the junction be T. K1 : K2 = 2 : 3 Q K1A 100 T K 2 A T 0 t L L k1 k2 l l 2 K1 K 2 (c) 2 K ML T 3 K –1 43. 200 – 2T = 3T 5T = 200 T = 40°C In steady state (a) temperature does not change with time (b) all parts of the body are at same temperature (c) there is no flow of heat (d) all of the above (a) - In steady state, temperature of the region is constant w.r.t. to time. - The temperature may be different for different parts of same body. - The heat flow will not be zero but equal in both direction. Two metallic plates of equal thicknesses and thermal conductivities K1 and K2 are put together face to face and a common plate is constructed, figure. The equivalent thermal conductivity will be: Ans. 2K1 K 2 (b) K K 1 (d) 2 K K 2 1 2K1 K 2 (b) Sol. l R1 K1A Req = R1 + R2 2l l l K eq A K1 A K 2 A 2 l l K eq K1 K 2 K eq 2K1K 2 K1 K 2 l K2A 2 3/ 2 2 R2 10 THERMAL PHYSICS 46. Two metallic plates of equal lengths and thermal conductivities k1 and k2 are put together such that their ends coincide. If their cross-sectional areas are the same, then the equivalent thermal conductivity of the combination will be : k1 k 2 (a) k k 1 k1 A k2 2 k1 k 2 2 (c) Ans. A 2 – 100 312.5C Temperature gradient on 100°C side 2k1k 2 (b) k k 1 (d) 2 25l 25 1 2KA 2 400 10 –4 206.25 –100 12 212.5 C m 48. Three rods made of same material and having same cross - section have been joind as shown in the figure. Each rod is of same length. The left end is kept at 0°C and both the right ends are kept at 90°C. The temperature of the junction of the three rods (in degree Celsius) will be Ans. 60.00 Sol. Let the junction temperature be in equilibrium k1 k 2 (c) Sol. 1 1 1 R e q R1 R 2 2k eq A l k1A k 2 A l l dQ dQ dQ dt rod1 dt rod 2 dt rod 3 2 K eq k1 k 2 Ans. k1 k 2 2 The ends of a copper rod of length 1m and area of cross 2 section 1 cm are maintained at 0°C and 100°C. At the centre, power is supplied at a constant rate of 25 J/s. The temperature gradient on higher temperature side of the rod in steady state (in °C/m) will be (K = 400 J/m-K-s) 212.50 Sol. Let be the temperature at the middle in steady state. K eq 47. 49. dQ dQ Then, P dt dt 1 KA 0 – 2KA 2 – 100 I Ans. KA 90 – KA 90 – 60 Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity K and other 2K. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport certain amount of heat from the hot end in the configuration I. The time (in seconds) to transport the same amount of heat in the configuration II is: KA KA or 25 1 2 –100 1 2 – 0 2.00 11 THERMAL PHYSICS Sol. 51. Q T i (R = thermal resistance) t R Q t T R ...(1) Since Q and T are the same t constant. Hence R The temperature of the two outer surface (end surfaces shown) of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K, thickness ‘x’ and ‘4x’ are T2 and T1(T2 > T1). The rate of heat transfer through the slab in steady state is A T2 – T1 K 1 , where ‘f’ is equal to x f t1 T 2 R1 R 2 (all quantities measured in S.I unit) Configuration 1: R 1 1 1 3 KA 2KA 3KA 1 2KA KA 3KA Configuration 2: R 2 3KA Using these in (i) R2 50. t 9 2 t 2s 3 2KA 3KA Figure shows a metal rod of uniform cross section area A, with variable thermal conductivity given by k x k 0 sec x . If the end A is maintained at 6L Ans. Sol. 3.00 Let temperature of common interface be T°C. Rate of heat flow H Q KAT t Q 2KA T – Ti H1 4x t 1 temperature T0 , the rod carries a thermal current I0 (from KA T2 – T Q and H 2 x t 2 I0 L B to A) in steady state and k AT 3 ; find the 0 1 In steady state, the rate of heat flow should be same in whole system i.e. temperature of the end B of the rod. Let’s say this temperature is kT0, find integer value k. H1 H 2 or 2KA T – Ti 4x KA T2 – T x T – Ti T2 – T 2 or T – T1 2T2 – 2T Ans. Sol. or T 2.00 I0 k0A dT dx cos x 6L 6I L T T0 0 sin x k 0 A 6L T 2T0 at x L 2T2 T1 3 ...(1) Hence heat flow from composite slab is H KA T2 – T x KA T2 – T1 3x 2T T1 KA T2 – 2 x 3 ...(2) 12 THERMAL PHYSICS 52. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperature T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2 = _______. Sol. Ans. Thermal radiations, are electromagnetic waves Thus radiation travels with speed of light. Heat is transferred most rapidly by the process of (a) Conduction (b) Convection (c) Radiation (d) Combustion (c) Sol. Heat radiation T 4 i.e. Fastest. 56. Two circular discs A and B with equal radii are blackened. They are heated to same temperature and then cooled under identical conditions. What inference do you draw from their cooling curves as shown in figure? 55. B A R 4.00 Since rate of heat flow is same, we can say ( – 0) K1 A 2 4 K 2 A1 Radiation 53. A sphere, a cube and a thin circular plate, all of same material and same mass, are initially heated to same high temperature. Choose the correct statement. (a) The plate will cool fastest and cube the slowest. (b) The sphere will cool fastest and cube the slowest. (c) The plate will cool fastest and sphere the slowest. (d) The cube will cool fastest and plate the slowest. Ans. (c) Sol. For same mass and density, surface area of plate will be largest and surface area of sphere will be smallest. The more surface area the more is rate of cooling. 54. Velocity of heat radiation v as related to the velocity of light c is (a) v > c (b) v = c (c) v < c (d) no definite relation Ans. (b) Ans. Sol. 57. (a) (b) Temperature L1 L2 K1A1 K 2 A 2 Time (c) (d) Time Ans. Sol. (c) In convection dQ T2 T1 dt dT Time 1 t Temperature R1 R 2 (a) A and B have same specific heats (b) specific heat of A is less (c) specific heat of B is less (d) nothing can be said (c) As the slope of A is less than slope of B. so rate of cooling will be slow in A so it has larger specific heat The temperature of coffee in a cup with time is most likely given by the curve in figure. Temperature 300 – 200 200 –100 R1 R2 Temperature Ans. Sol. Time 13 THERMAL PHYSICS A block of steel heated to 100°C is left in a room to cool. Which of the curves shown in the figure represents the decrease of temperature with time? Temperature 58. B Sol. dT K T T0 dt D C K A 62. Ans. (a) A (c) C (a) Sol. dT 59. A body cools from 50°C to 46°C in 5 minutes and to 40°C in the next 10 minutes. The surrounding temperature is : (a) 30°C (b) 28°C (c) 36°C (d) 32°C (b) Sol. (b) B (d) D 1 t dT / dt 0.2C / min 0.01 / min T T0 20C Ans. Newton’s law of cooling is applicable for (a) Any excess of temperature over the surrounding (b) Small excess of temperature over the surrounding (c) Large excess of temperature over the surrounding (d) Very large excess of temperature over the surrounding (b) Newton’s law of cooling of only for small excess of temperature otherwise value of K will change law Newton’s law of cooling leads us to the following expression, (a) ( – 0) = Kt + C (b) log ( – 0) = Kt + C (c) log = Kt + C (d) = K0 + C (b) Sol. From Time Ans. dT K T T0 dt Ans. Sol. 63. msdT1 K TAV1 TS t1 dQ K 0 dt dQ K d t 0 ms 4 K 48 TS 5 64. msdT2 K TAV 2 TS t2 ms 6 K 43 TS 10 60. Ans. Sol. 61. Ans. log 0 Kt C ...(i) Ans. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9 (a) Sol. E AT 4 ...(ii) TS = 28°C Newton’s law of cooling is used in laboratory for determining: (a) Specific heat of gases (b) Specific heat of liquids (c) Latent heat of gases (d) Latent heat of liquids (b) Newton’s law of cooling is used to find specific heat of liquids. If the rate of change of temperature is 0.2°C/ min and excess temperature of a body over surrounding is 20°C, the constant of proportionality is (a) 0.1 (b) 0.01 (c) 1 (d) 0.001 (b) E1 4R12 T14 1 4000 1:1 E 2 4R 22 T24 4 2 2000 4 2 65. Ans. Sol. 66. 4 Which of the following is more close to a black body ? (b) Black board paint (b) Green leaves (c) Black holes (d) Red roses (c) Black hole is black body. Infrared radiations are detected by (a) spectrometer (b) pyrometer (c) nanometer (d) photometer 14 THERMAL PHYSICS Ans. Sol. 67. (b) Infrared radiations are detected by pyrometer. A liquid in a beaker has temperature (t) at time t and 0 is temperature of surroundings, then according to Newton’s law of cooling, the correct graph between loge ( – 0) and t is : Sol. 70. Time of cooling increase with decrease in temperature. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on earth, at a distance r from the sun, where r0 is radius of earth (a) 4r02 R 2 T 4 r2 (b) r02 R 2 T 4 r2 (c) r02 R 2 T 4 4r 2 (d) R 2 T 4 r2 Ans. (b) Sol. Power emitted by sun AT 4 4R 2 T 4 Energy per area per unit time at earth 4R 2 T 4 R 2 T 4 4r 2 r2 Energy power incident on earth. 71. Ans. Sol. R 2 T 4 r02 r2 A piece of metal is heated to temperature and then allowed to cool in a room which is at temperature 0. The graph between the temperature T of the metal and time t will be closed to : T (d) According to Newton’s law of cooling (a) T O In 0 kt C 68. So, graph in option - (d) is correct. The maximum wavelength of radiations emitted at 900 K is 4 m. What will be the maximum wavelength of radiations emitted at 1200K: (a) 3 m (b) 0.3 m Ans. (c) l m (a) Sol. 1T1 2 T2 (d) none of these 4m 900 2 1200 2 3 m 69. Ans. A bucket full of hot water is kept in a room and it cools from 75°C to 70°C in T1 minutes, from 70°C to 65°C in T2 minutes and from 65°C to 60°C in T3 minutes then (a) T1 = T2 = T3 (b) T1 < T2 > T3 (c) T1 > T2 > T3 (d) T1 < T2 < T3 (d) (b) t O T T (c) (d) O Ans. Sol. t t O t (c) According to Newton’s law of coding temperature decrease exponentially with time and temperature becomes equal to room temp. at infinity time. 15 THERMAL PHYSICS 72. Ans. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously, will be (a) 4 (b) 16 (c) 32 (d) 64 (d) 4R T 2 Sol. Energy received by earth 4r 2 4 75. Ans. Sol. T K T – T0 t r t 2 0 For cooling from 60°c to 40°C So, E R 2 T 4 E f 2R 2T 64 So, EL R 2T4 2 73. A body cools in 7 minutes from 60°C. What will be its temperature (in °C) after the next 7 minutes? The temperature of surroundings in 10°C. 28.00 Rate of cooling 60 – 40 20 1 60 40 K –10 K 7 7 40 14 2 For cooling from 40°C to T 4 In the figure, the distribution of energy density of the radiation emitted by a black body at a given temperature is shown. The possible temperature of the black body is 3 n × 10 K. Write n as the nearest integer. –3 (Take:- b = 2.898 × 10 mK) 40 – T 40 – T 1 40 T – 20 40 T K –10 7 7 14 2 2 160 – 4T 20 T 140 5T T 140 23C 5 T 28C 76. A solid cube of side a, density d and specific heat ‘s’ is at temperature 400 K. It is placed in an ambient temperature of 200 K. –3 3 3 Take: a = 0.9 m, d = 4.8 ×10 kg/m , s = 2.0 × 10 J/kg K. Stefan’s constant 6 10 –8 W K 4 m 2 . Consider the Ans. 2.00 Ans. cube to be a black body. If the time for the temperature of the cube to drop by 5 K is 1000x second, find x in nearest integer. 5.00 Sol. m T b where b 2.898 10 –3 mK Sol. T 4 – TS4 6a 2 t d.a 3 s.T T 74. b 2.898 10–3 1932K 1.9 103 K m 1.5 10–6 So, nearest integer value of n will be 2. Two starts (say S1 and S2) emit their maximum intensity –6 –7 radiations at wavelengths 10 m and 0.4 × 10 m t 77. T2 respectively. The ratio of their surface temperature T 1 Ans. Sol. is 25.00 T 10 –6 m T constant 2 1 25 T1 2 0.4 10 –7 Ans. Sol. dos T 6. T 4 – Ts4 4.8 103 0.9 2.0 103 5 6 6 10 –8 400 4 – 200 4 5000s. x5 Two planets X and Y revolving around the sun is circular orbits, have temperature of their surfaces as T1 and T2. If their distance from the sun are in the ratio of 1 : 4, then find T1 : T2. Assume the planets to be in the steady state, and the sun and the planets to be black-bodies. Neglect the energy exchange between the two planets. 2.00 dQsum eA s T 4 dt dQ rod. dt dQsum dt 4d 2 R 2p 16 THERMAL PHYSICS sol. dQ rod. In steady state dt From Newton’s law of cooling dT k T – Tsurrounding dt dQ radiated 4R 2p tP 2 dt ...(1) Integration equation (1) Tf – Tsurrounding 14 As Tp T 2 2 16 d Ti – Tsurrounding e – kt ...(2) From given data in question in first 7 minutes 1 2 p Tp d constant. Tf 40C, Ti 60C, Tsurrounding 10C T1 : T2 2: 1 78. A glass of boiling water at 100°C cools down to 90°C in 10 minutes when placed in surrounding temperature of 30°C. It will cool down to 80°C in an additional time t 10 ...(3) Now the temperature decrease further due to temperature gradient from given data in question in next 7 minutes nx – n x –1 Tf ?, Ti – 40C, Tsurrounding 10C n x 1 – nx minutes. Assume Newton’s law of Ans. cooling to hold. Find x 6.00 Sol. T – T0 Ti – T0 e – kt 60 70e –10k k applying equation (1) Tt – 10 e– k 7 40 – 10 1 7 n 10 6 Tf –10 3 40 – 10 5 T – T0 Ti – T0 e – kt 50 60e – kt 6 6 kt n t 10 5 7 5 n 6 79. Ans. A body cools in 7 minutes from 60°C to 40°C. What will be its temperature in °C after next 7 minutes? The temperature of surrounding is 10°C. 28.00 ...(4) substituting value of e – k 7 , from equation (3) For further cooling, n 40 – 10 e– k 7 60 – 10 Tf 10 18 28C 80. Ans. Sol. The emissive power of a black body at T = 300 K is 460 W/m2. Consider a body B of area A = 100 cm2 , coefficient of reflectivity r = 0.3 and coefficient of transmission t = 0.5. Its temperature is 300 K. Then the power radiated by B is ____ W. 0.92 Since, e = a = 0.2 (Since, a = (1–r–t) = 0.2 for the body B) E = (460) (0.2) = 92 W/m2 P = EA = 92 × 100 × 10–4 = 0.92W