2022 thermo Q1: a) i) Thermodynamic system: It refers to the part of the universe under study which is separated from the rest of the universe by a system boundary. It may contain matter or radiation, and may undergo changes as a result of interactions with the surroundings. ii) System boundary: It is the boundary that separates the thermodynamic system from the surroundings. b) i) Closed system: A closed system is one which allows energy transfer but not mass transfer across its boundary. ii) Open system: An open system is one which allows both mass and energy transfer across its boundary. iii) Isolated system: An isolated system is one which does not allow either mass or energy transfer across its boundary. d) Entropy is a measure of the degree of disorder in a system. It is found to increase with the temperature of the working substance. At low temperatures, the increase in entropy is less rapid, whereas at high temperatures, the increase in entropy is more rapid. e) i) Isochoric process: In an isochoric process, the volume of the system remains constant, i.e. there is no work done. The process takes place at a constant volume, and the pressure and temperature of the system may change. ii) Isobaric process: In an isobaric process, the pressure of the system remains constant, i.e. the process takes place at a constant pressure, and the volume and temperature of the system may change. iii) Isothermal process: In an isothermal process, the temperature of the system remains constant, i.e. the process takes place at a constant temperature, and the pressure and volume of the system may change. iv) Adiabatic process: In an adiabatic process, there is no heat transfer between the system and the surroundings. The process takes place without any exchange of heat, and the pressure and volume of the system may change. Q2. a) The first law of thermodynamics is the law of energy conservation, which states that energy can neither be created nor destroyed, but it can be converted from one form to another. b) [The poster can vary based on personal preference and style, but here are some ideas for practical applications of thermodynamics:] • • • • Power generation: Thermodynamics is used to generate electricity in power plants by converting heat energy from combustion, nuclear fission, or geothermal sources into mechanical energy and then to electrical energy. HVAC systems: Heating, ventilation, and air conditioning (HVAC) systems use thermodynamics principles to control temperature, humidity, and air quality in buildings, homes, and vehicles. Engines and turbines: Thermodynamics principles are used to design and optimize internal combustion engines, gas turbines, jet engines, and other types of engines and turbines for various applications such as transportation, power generation, and aviation. Refrigeration and cooling: Thermodynamics principles are used to design and optimize refrigeration and cooling systems for various applications such as food preservation, air conditioning, and medical equipment. c) To calculate the net work done on the oxygen gas during the process, we can use the following steps: • • • Calculate the initial internal energy of the oxygen gas: Since the gas is at a known pressure and temperature, we can use the ideal gas law to find its initial specific volume, which is v1 = RT1/P1, where R is the gas constant, T1 is the initial temperature, and P1 is the initial pressure. Then we can find the initial internal energy per unit mass of the gas as u1 = CvT1, where Cv is the specific heat at constant volume. Calculate the final internal energy of the oxygen gas: Since the gas is adiabatically expanded, its specific entropy remains constant, which means the final temperature can be found using the relation P2v2^k = P1v1^k, where k is the polytropic index. Then we can find the final internal energy per unit mass of the gas as u2 = CvT2. Similarly, since the gas is compressed at constant pressure and then compressed again at constant volume, we can find the internal energy at each state using the appropriate specific heat and temperature. Calculate the net work done on the oxygen gas: The net work done on the gas is equal to the sum of the work done during each process, which can be calculated using the relation W = ∫PdV, where the integration is performed over the appropriate volume range for each process. Since the gas is expanding and then compressing, the net work done on the gas is negative, which means work is done on the surroundings. The final answer can be converted to joules by multiplying by the mass of the gas. d) Intensive properties of a substance are those that do not depend on the amount of the substance present. Some examples of intensive properties are: • • • • • • • • Temperature Pressure Density Specific heat Enthalpy Entropy Internal energy Gibbs free energy Q3. a) Mechanical work is the work done by a force on an object through a distance, while thermodynamic work is the work done by or on a thermodynamic system due to a change in volume or pressure. b) Given: pressure, P = 200 kPa, current, I = 0.70 A, voltage, V = 110 V, m = 300 g, time, t = 10 min = 600 s, heat loss, Q_loss = 2.2 kJ. Using the first law of thermodynamics, the work done during the process can be calculated as: Q - W = ΔU where Q is the heat added to the system, W is the work done by the system, ΔU is the change in internal energy. The heat added to the system is: Q = m * (hf + Cp * ΔT) where hf is the enthalpy of vaporization, Cp is the specific heat capacity, and ΔT is the change in temperature. hf for water at 200 kPa can be found from the steam tables to be 2416 kJ/kg, and Cp for water can be taken as 4.18 kJ/kg·K. The current, voltage, and time can be used to calculate the energy supplied to the system as: E=V*I*t The change in internal energy can be assumed to be negligible for this process, since the piston-cylinder device is not changing in volume. Therefore, the work done by the system is: W = Q - Q_loss = m * (hf + Cp * ΔT) - Q_loss The final temperature of the steam can be found by rearranging the equation for Q and solving for ΔT: ΔT = (Q - m * hf) / (m * Cp) The p-v diagram of the Carnot power cycle is a rectangle, representing two isothermal and two adiabatic processes. The heat is added to the system during the isothermal expansion, and rejected during the isothermal compression, while the temperature difference between the hot and cold reservoirs drives the adiabatic expansion and compression. Q4. a) The three alternative statements of the second law of thermodynamics are: 1. The Kelvin-Planck statement: It is impossible to construct a heat engine that produces no effect other than the extraction of heat from a hot reservoir and the performance of an equivalent amount of work. 2. The Clausius statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cold body to a hot body. 3. The entropy statement: For a closed system, the entropy never decreases with time. b) A reversible process is a process that can be reversed at every stage, by making infinitesimal changes to its state variables. It occurs very slowly, so the system is always in thermodynamic equilibrium with its surroundings. In contrast, an irreversible process is a process that cannot be reversed to its initial state. It occurs quickly, and the system may not be in thermodynamic equilibrium at every stage. c) The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir to the work input required to run the refrigerator. Mathematically, COP = Qc / Win, where Qc is the rate of heat transfer to the refrigerant from the freezer compartment and Win is the power input required to operate the refrigerator. COP = Qc / Win = (8000 kJ/h) / (3200 kJ/h) = 2.5 The coefficient of performance of a reversible refrigeration cycle operating between reservoirs at the same two temperatures is given by the Carnot efficiency formula: COP_reversible = Th / (Th - Tc) = (22 + 273) / [(22 + 273) - (-5 + 273)] = 8.18 Therefore, the COP of the actual refrigerator is lower than the COP of the reversible refrigeration cycle. Q5. a) A power cycle is a thermodynamic cycle in which heat is converted into work. Two examples of real-world systems with power cycles are: 1. Gas turbine power plant: In a gas turbine power plant, air is compressed and mixed with fuel which is then burned, generating high temperature and high pressure gases. These gases expand and pass through a turbine, producing mechanical work that is used to drive a generator to produce electricity. 2. Internal combustion engine: In an internal combustion engine, a fuel-air mixture is ignited within a combustion chamber, generating high temperature and high pressure gases. These gases expand, creating pressure that moves a piston, producing mechanical work that is used to propel a vehicle. b) A household refrigerator works on the principle of a vapor-compression refrigeration cycle. The working principle of a refrigerator can be explained as follows: • • • • • The refrigerant (usually a gas) is compressed by the compressor, which raises its pressure and temperature. The high-pressure, high-temperature refrigerant flows through the condenser coils on the back of the refrigerator, where it releases heat to the surrounding air and condenses into a liquid. The liquid refrigerant flows through the expansion valve, which reduces its pressure and temperature. The low-pressure, low-temperature refrigerant flows through the evaporator coils inside the refrigerator, where it absorbs heat from the food and air inside the refrigerator and vaporizes into a gas. The gas refrigerant is then drawn back into the compressor, where the cycle begins again. The overall effect of this cycle is to transfer heat from inside the refrigerator to the outside, thus keeping the inside cool. A sketch of the refrigerator working principle is shown below: c) The rate of heat transfer to the ocean by the ship's heat engine can be calculated as follows: • • • The power output of the engine is given as 75000 kW. The thermal energy released by the fuel oil burned per hour is given as 13600 litres x 43000 kJ/litre = 585.8 x 10^6 J/h. The rate of heat transfer to the ocean can be calculated as the difference between the thermal energy released by the fuel oil and the power output of the engine, or (585.8 x 10^6 J/h) - (75000 kW x 3600 s/h) = 314.8 x 10^6 J/h. The thermal efficiency of the ship's heat engine can be calculated as the ratio of the power output to the thermal energy released by the fuel oil burned, or (75000 kW)/(585.8 x 10^6 J/h) = 0.128, or 12.8%.
