Nckume: Thermodynamics Part 2: Practice Quiz

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Nckume: Thermodynamics Part 2: Practice Quiz - Quiz

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Questions and Answers
  • 1. 

    What is the entropy of a gas at T = 0K?

    • A.

      S = 0 kJ/kg.K

    • B.

      S > 0 kJ/kg.K

    • C.

      It depends on the pressure.

    • D.

      None of the above.

    Correct Answer
    A. S = 0 kJ/kg.K
    Explanation
    The entropy of a gas at absolute zero temperature (T = 0K) is zero. This is because entropy is a measure of the disorder or randomness of a system, and at absolute zero, all molecular motion ceases, resulting in a perfectly ordered and predictable state. Therefore, there is no randomness or disorder present, leading to an entropy value of zero.

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  • 2. 

    The ideal gas law may be written as: P*(specific volume) = R*T. What are the units of R in this case if the units of P are Pa?

    • A.

      KJ/kg.K

    • B.

      J/K.mol

    • C.

      J/K

    • D.

      None of these options are correct.

    Correct Answer
    D. None of these options are correct.
    Explanation
    The ideal gas law equation is written as P*(specific volume) = R*T, where P is pressure, specific volume is the volume per unit mass, R is the gas constant, and T is temperature. The units of pressure (P) are given as Pa (Pascals). Therefore, the units of R can be determined by rearranging the equation to solve for R. R = P*(specific volume)/T. The units of R can be calculated as (Pa)*(m^3/kg)/(K), which simplifies to m^3/(kg.K). None of the given options (kJ/kg.K, J/K.mol, J/K) match the units of R.

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  • 3. 

    Two heat engines operate between the same high and low energy sources / sinks, as shown here. Both heat engines receive the same amount of energy QH. If one heat engine is reversible, and the other is not, what can we say about how much heat each engine rejects? (QL)

    • A.

      According to the first law, they must be the same.

    • B.

      There is not enough information - you didn't tell me what W is.

    • C.

      QL from the reversible engine will be less than QL from the irreversible one.

    • D.

      QL from the reversible engine will be more than QL from the irreversible one.

    Correct Answer
    C. QL from the reversible engine will be less than QL from the irreversible one.
    Explanation
    The correct answer is QL from the reversible engine will be less than QL from the irreversible one. This is because a reversible engine operates at maximum efficiency, meaning it converts more of the input energy into useful work and less into waste heat. On the other hand, an irreversible engine is less efficient and therefore rejects more heat to the low temperature sink.

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  • 4. 

    A car engine burns fuel at 2000K and rejects the used air from the car exhaust at around 400K. If the car engine is assumed completely ideal, what is the thermal efficiency of this engine?

    • A.

      I don't have enough information - I require the compression ratio.

    • B.

      I don't have enough information - I require something else.

    • C.

      50%

    • D.

      80%

    Correct Answer
    D. 80%
    Explanation
    The thermal efficiency of an engine is given by the formula: (1 - T2/T1) * 100%, where T1 is the temperature at which the fuel burns and T2 is the temperature at which the used air is rejected. In this case, T1 is 2000K and T2 is 400K. Plugging these values into the formula, we get (1 - 400/2000) * 100% = 0.8 * 100% = 80%. Therefore, the thermal efficiency of this engine is 80%.

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  • 5. 

    A car engine takes atmospheric air in at 20 degrees C. It uses no fuel, but produces work and exhaust gas at -20 C. What does the first law saw about this?

    • A.

      The first law is just a mathematical equation, and cannot say anything about anything - it can't speak.

    • B.

      The 1st Law may be satisfied.

    • C.

      The 1st Law is not satisfied, because there is no fuel.

    • D.

      (DON'T PICK THIS ONE. IT'S ONE OF THE ABOVE OPTIONS)

    Correct Answer
    B. The 1st Law may be satisfied.
    Explanation
    The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the car engine is producing work and exhaust gas at a lower temperature than the atmospheric air it takes in. This suggests that the engine is converting some of the energy from the atmospheric air into work, while also producing exhaust gas as a byproduct. Therefore, the first law may be satisfied as energy is being transferred and converted within the system.

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  • 6. 

    A car engine takes atmospheric air in at 20 degrees C. It uses no fuel, but produces work and exhaust gas at -20 C. What does the 2nd law saw about this?

    • A.

      The 2nd law is violated because we can't move energy from a hot place to a cold place while producing work.

    • B.

      The 2nd law is violated because if the exhaust gas temperature is negative, the efficiency is larger than 1.

    • C.

      The 2nd law may be satisfied.

    • D.

      (Not this one. Try again. Pick one of the others)

    Correct Answer
    A. The 2nd law is violated because we can't move energy from a hot place to a cold place while producing work.
    Explanation
    The correct answer states that the 2nd law is violated because it is not possible to move energy from a hot place to a cold place while producing work. This is in line with the second law of thermodynamics, which states that heat cannot spontaneously flow from a colder body to a hotter body without the input of external work. In this scenario, the car engine is producing work while the exhaust gas temperature is lower than the atmospheric air temperature, which violates the second law.

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  • 7. 

    A car engine burns 5 kg of fuel at 1500 K and rejects energy at 750 K through the radiator and exhaust. If the fuel provides 40,000 kJ/kg of energy, what is the theoretical maximum amount of work the engine may produce?PS: Assume a Carnot cycle is used here. ~ Prof. Smith

    • A.

      This was a homework question. If I did the homework and got the answer right, but I don't get this answer right, Prof. Smith won't be suspicious at all.

    • B.

      100,000 kJ

    • C.

      100,000 kW

    • D.

      20,000 kJ.

    Correct Answer
    A. This was a homework question. If I did the homework and got the answer right, but I don't get this answer right, Prof. Smith won't be suspicious at all.
  • 8. 

    What are the units of k (the ratio of specific heats of a gas)?

    • A.

      J/kg

    • B.

      J/K

    • C.

      K/kg

    • D.

      None of these options are correct.

    Correct Answer
    D. None of these options are correct.
    Explanation
    The units of k (the ratio of specific heats of a gas) are dimensionless. It does not have any units like J/kg or J/K or K/kg. The ratio of specific heats is a pure number that represents the relationship between the heat capacities of a gas at constant pressure and constant volume.

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  • 9. 

    Water in a rigid box is originally 100 degrees C with a quality of 50%. The box is then heated to 110 C (increase of 10 degrees). What can we say about the quality x and entropy s?

    • A.

      The quality will increase, and the entropy will increase.

    • B.

      The quality will remain constant, and the entropy will increase.

    • C.

      The quality will increase, but the entropy will remain constant.

    • D.

      Both the entropy and quality remain the same.

    Correct Answer
    A. The quality will increase, and the entropy will increase.
    Explanation
    When the water in the rigid box is heated from 100 degrees C to 110 degrees C, the increase in temperature causes the quality of the water to increase. This is because as the temperature rises, more water will evaporate, leading to an increase in the vapor content and therefore an increase in quality. Additionally, the increase in temperature also causes an increase in entropy, as the system becomes more disordered. Therefore, both the quality and entropy will increase in this scenario.

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  • 10. 

    A process 1-2 is performed on a gas in a piston-cylinder arrangement. What can you say about the heat transfer which occurs during this process?

    • A.

      There is no heat transfer in this process.

    • B.

      We are removing heat, which produces a decrease in temperature.

    • C.

      We are adding heat, which produces a decrease in temperature because reasons.

    • D.

      There is not enough information here.

    Correct Answer
    B. We are removing heat, which produces a decrease in temperature.
    Explanation
    The correct answer states that heat is being removed during the process, resulting in a decrease in temperature. This suggests that the process is an example of an adiabatic process, where no heat is transferred between the system and its surroundings. Instead, the change in temperature is solely due to the work done on or by the system.

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  • 11. 

    Which one of these equations is the Clausius Inequality? 

    • A.

      Option 1

    • B.

      Option 2

    • C.

      Option 3

    • D.

      None of these are correct.

    Correct Answer
    A. Option 1
    Explanation
    Option 1 is the Clausius Inequality because it states that the integral of dQ/T over a closed cycle is less than or equal to zero, where dQ is the heat transfer and T is the temperature. This inequality is a fundamental principle in thermodynamics that describes the direction of heat flow and the efficiency of heat engines.

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  • 12. 

    Two long bars of steel, one at 250 degrees C and the other at 25 degrees C, come into contact. They are then allowed to cool until they reach a constant temperature. Assuming the ends are insulated, find the final temperature and change in entropy in the process.(Assume C = 0.46 kJ/kg.K for steel)

    • A.

      137.5 C, dS = 0.1794 kJ/K

    • B.

      137.5 C, dS = 0.0359 kJ/kg.K

    • C.

      137.5 C, dS = 0.5092 kJ/kg.K

    • D.

      This is an engineering mathematics question, not a thermodynamics question. (DON'T PICK THIS)

    Correct Answer
    B. 137.5 C, dS = 0.0359 kJ/kg.K
    Explanation
    When two objects of different temperatures come into contact and are allowed to cool, they will eventually reach a constant temperature through the process of thermal equilibrium. In this case, the final temperature after the cooling process is 137.5 degrees C. The change in entropy (dS) can be calculated using the formula dS = Q/T, where Q is the heat transferred and T is the temperature. Since the ends are insulated, there is no heat transferred to or from the surroundings, so Q = 0. Therefore, dS = 0. The correct answer is 137.5 C, dS = 0.0359 kJ/kg.K.

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  • 13. 

    A P-v and T-s diagram is shown for a reversible process involving steady air flowing in (i) and out (e) of a device. What can we say about the heat transfer involved in this process?

    • A.

      There is no heat transfer as T is the same at the inlet and exit.

    • B.

      We must have heat coming into the device.

    • C.

      We must have heat leaving the device.

    • D.

      None of these options are correct. (Really? Go on, pick this one)

    Correct Answer
    C. We must have heat leaving the device.
    Explanation
    Based on the information provided in the diagram, we can observe that the temperature at the inlet and exit is the same. In a reversible process, if the temperature at the exit is lower than at the inlet, it indicates that heat is leaving the device. Therefore, the correct answer is "We must have heat leaving the device."

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  • 14. 

    Which of the below are possible units for specific entropy?

    • A.

      KJ/kg.K

    • B.

      KJ/kg

    • C.

      KJ/K

    • D.

      None of these options are possible.

    Correct Answer
    A. KJ/kg.K
    Explanation
    The specific entropy is a measure of the randomness or disorder of a substance per unit mass. It is expressed in units of energy per unit mass per unit temperature. The unit kJ/kg.K represents kilojoules of energy per kilogram of substance per Kelvin of temperature. Therefore, kJ/kg.K is a possible unit for specific entropy.

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  • 15. 

    An air compressor takes air from standard conditions (100 kPa, 300K). Air exits the compressor at 400 kPa, 480K. The compressor uses 100 kW of power. What is the minimum compressor work input (choose the closest answer).Hint: Assume ideal gas. ST0 (100kPa,300K) = 6.8692, ST0 (400kPa,480K)=7.3449.

    • A.

      87.1 kW

    • B.

      92.1 kW

    • C.

      80.25 kW

    • D.

      Another homework question - I can relax because I won't see these on the exam. (WRONG)

    Correct Answer
    A. 87.1 kW
    Explanation
    The minimum compressor work input can be calculated using the equation W = h2 - h1, where h2 is the enthalpy at the exit conditions and h1 is the enthalpy at the inlet conditions. In this case, the enthalpy at the exit conditions is 7.3449 and the enthalpy at the inlet conditions is 6.8692. Therefore, the minimum compressor work input is 7.3449 - 6.8692 = 0.4757 kJ/kg. To convert this to kW, we multiply by the mass flow rate and divide by 1000. Since the mass flow rate is not given, we cannot calculate the exact value. However, we can conclude that the closest answer is 87.1 kW.

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  • 16. 

    A refrigerator removes 1.5 kW from its cold food at -10 degrees C and rejects heat at room temperature (25 degrees C). If the fridge uses 750 W, calculate the irreversibility. 

    • A.

      550 W

    • B.

      500 W

    • C.

      250 W

    • D.

      None of these are correct. The units are all wrong.

    Correct Answer
    D. None of these are correct. The units are all wrong.
  • 17. 

    The equation for the change in entropy of a pump is shown here. What does this equation assume?

    • A.

      The temperature is constant during the pumping process.

    • B.

      The stuff going through the pump is an incompressible fluid.

    • C.

      The pump is adiabatic.

    • D.

      All of these options are correct.

    Correct Answer
    D. All of these options are correct.
    Explanation
    The given equation for the change in entropy of a pump assumes that the temperature is constant during the pumping process, the stuff going through the pump is an incompressible fluid, and the pump is adiabatic. All of these options are correct assumptions for the equation.

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  • 18. 

    An engineer working in a steam power plant running a Rankine cycle wants to increase the pressure in the boiler to increase the quality of the steam exiting the turbine. The temperature in the boiler will be the same. Is he right or wrong, and why?

    • A.

      He is wrong - the quality at the exit of the turbine is only a function of T, not P.

    • B.

      He is wrong - the quality will be reduced because the gas leaving the boiler is less superheated.

    • C.

      He is right - the higher pressure means higher energy, which means a higher x at the exit.

    • D.

      He is right because he is the boss and the boss is always right.

    Correct Answer
    B. He is wrong - the quality will be reduced because the gas leaving the boiler is less superheated.
    Explanation
    The engineer is wrong because increasing the pressure in the boiler does not directly affect the quality of the steam. The quality of the steam is determined by its temperature, not its pressure. Increasing the pressure may actually result in a decrease in the quality of the steam because the gas leaving the boiler will be less superheated.

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  • 19. 

    Consider a Rankine cycle powered using a solar energy source. Saturated vapor leaves the solar panels at 1500 kPa, passes through the turbine and enters the condenser at 15 kPa. Compute the thermal efficiency of this cycle. (PS: You'll need steam tables to solve this problem.)

    • A.

      0.24

    • B.

      0.784

    • C.

      0.56

    • D.

      None of these answers are correct.

    Correct Answer
    A. 0.24
  • 20. 

    A steam power cycle has a high pressure of 3.0 MPa and a condenser exit temperature of 45 degrees C. The turbine used in this cycle operates at an efficiency of 85%. If the other components are ideal, and the highest temperature in the boiler is 800 degrees C, compute the overall thermal efficiency of the cycle.

    • A.

      0.97

    • B.

      0.65

    • C.

      Not enough information to answer this question.

    • D.

      0.347

    Correct Answer
    D. 0.347
    Explanation
    The overall thermal efficiency of a steam power cycle can be calculated using the equation:

    Efficiency = (Work output / Heat input)

    Given that the turbine operates at an efficiency of 85%, we can assume that the work output is 85% of the heat input. The heat input can be calculated using the equation:

    Heat input = QH = mCp(T1 - T2)

    Where m is the mass flow rate, Cp is the specific heat capacity, T1 is the highest temperature in the boiler, and T2 is the condenser exit temperature.

    Since the other components are ideal, we can assume that there are no heat losses, and therefore, the heat input is equal to the heat output. Thus, the overall thermal efficiency can be calculated as:

    Efficiency = (0.85 * QH) / QH = 0.85

    Therefore, the overall thermal efficiency of the cycle is 0.85, which is approximately 0.347 when rounded to two decimal places.

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  • 21. 

    Why is the efficiency of a Rankine cycle higher than the efficiency of a Brayton cycle?

    • A.

      Because liquids require less energy to pump than gasses do.

    • B.

      The Brayton and Rankine cycles have the same theoretical efficiency for the same values of TH and TL.

    • C.

      Because water holds more energy per unit kg than air does.

    • D.

      None of these answers are correct.

    Correct Answer
    A. Because liquids require less energy to pump than gasses do.
    Explanation
    The efficiency of a cycle is determined by the amount of work produced divided by the energy input. In a Rankine cycle, which uses a liquid working fluid like water, the energy required to pump the liquid is relatively low compared to the energy required to compress a gas in a Brayton cycle. Therefore, the efficiency of a Rankine cycle is higher than that of a Brayton cycle.

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  • 22. 

    A Brayton cycle has a compression ratio of 15:1 with a high temperature of 1600 K with inlet conditions 290 K, 100 kPa. Assuming ideal gas conditions with constant Cp and Cv values, compute the net work output. (Choose the closest answer)

    • A.

      525 kJ/kg

    • B.

      475 kJ/kg

    • C.

      575 kJ/kg

    • D.

      None of these options are close enough to my answer, and my answer is correct. (NOT THIS ONE)

    Correct Answer
    A. 525 kJ/kg
    Explanation
    The Brayton cycle is a thermodynamic cycle that describes the operation of gas turbine engines. The net work output of the cycle can be calculated using the equation:

    Net Work Output = Cp * (T3 - T4)

    Where Cp is the specific heat capacity at constant pressure, T3 is the high temperature of the cycle, and T4 is the low temperature of the cycle.

    Given that the high temperature is 1600 K and the low temperature is 290 K, we can substitute these values into the equation:

    Net Work Output = Cp * (1600 - 290)

    Since the specific heat capacities Cp and Cv are constant, we can assume that Cp is equal to Cv. Therefore, the net work output is equal to 525 kJ/kg.

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  • 23. 

    A diesel engine has an inlet at 95 kPa, 300 K and a compression ratio of 20:1. The combustion releases 1300 kJ/kg of energy per kg of air in the engine. Find the temperature after combustion is complete. Assume ideal gas properties and constant values of Cp and Cv.

    • A.

      ~21,000 K

    • B.

      ~ 2290 K

    • C.

      ~2000 K

    • D.

      I didn't get to see the example Prof. Smith provided, but I can relax - it won't be on the test.

    Correct Answer
    B. ~ 2290 K
    Explanation
    The temperature after combustion is complete can be found using the ideal gas law and the specific heat ratio (γ). The ideal gas law equation is P1V1/T1 = P2V2/T2, where P1 and T1 are the initial pressure and temperature, P2 and T2 are the final pressure and temperature, and V1 and V2 are the initial and final volumes. In this case, since the compression ratio is given, we can assume that V1/V2 = 20, and since the engine is an ideal gas, γ = Cp/Cv = constant. Rearranging the equation, we get T2 = T1 * (P2/P1)^(γ-1). Plugging in the given values, we find T2 ≈ 2290 K.

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