# Thermodynamics Questions and Answers – Second Law of Thermodynamics

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This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Second Law of Thermodynamics”.

1. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output for this heat engine. a) 30 MW
b) 40 MW
c) 50 MW
d) 60 MW

Explanation: Net power output = 80 – 50 MW = 30 MW.

2. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the thermal efficiency for this heat engine. a) 47.5 %
b) 27.5 %
c) 37.5 %
d) none of the mentioned

Explanation: The thermal efficiency of heat engine = net work output / heat input
= 30/80 = 0.375 = 37.5 %.

3. A car engine with a power output of 50 kW has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 44,000 kJ/kg . a) 0.00273 kg/s
b) 0.00373 kg/s
c) 0.00473 kg/s
d) 0.00573 kg/s

Explanation: Q = 50/0.24 = 208.3 kW,
hence fuel consumption rate = 208.3 kW / 44000 kJ/kg = 0.00473 kg/s.
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4. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the coefficient of performance of the refrigerator. a) 4
b) 3
c) 2
d) 1

Explanation: COP = (360/2)(1/60) = 3.

5. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the rate of heat rejection to the room that houses the refrigerator. a) 450 kJ/min
b) 460 kJ/min
c) 470 kJ/min
d) 480 kJ/min

Explanation: Q = 360 + (2)(60/1) = 480 kJ/min.

6. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the power consumed by the heat pump. a) 32000 kJ/h
b) 33000 kJ/h
c) 34000 kJ/h
d) 35000 kJ/h

Explanation: W = Q/COP = 80000 kJ/h / 2.5 = 32000 kJ/h.

7. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the rate at which heat is absorbed from the cold outdoor air. a) 32000 kJ/h
b) 48000 kJ/h
c) 54000 kJ/h
d) 72000 kJ/h

Explanation: The rate at which heat is absorbed = 80000 – 32000 = 48000 kJ/h.

8. An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air-conditioner.
a) 1.09 kW
b) 1.19 kW
c) 1.29 kW
d) 1.39 kW

Explanation: Q = m*cp*(temperature change) = 20.08 kW
COP = (15+273)/(35-15) = 14.4
hence power needed = 20/14.4 = 1.39 kW.

9. A cyclic machine, as shown below, receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200kJ of work as output. Is this cycle reversible, irreversible, or impossible?
a) reversible
b) irreversible
c) impossible
d) none of the mentioned

Explanation: The Carnot efficiency = 1 – (400/1000) = 0.6 and real efficiency = (300/325) = 0.615 which is greater than the Carnot efficiency hence cycle is impossible.

10. In a cryogenic experiment you need to keep a container at -125°C although it gains 100 W due to heat transfer. What is the smallest motor you would need for a heat pump absorbing heat from the container and rejecting heat to the room at 20°C?
a) 97.84 kW
b) 98.84 kW
c) 99.84 kW
d) 95.84 kW

Explanation: COP = 1.022 and thus power required = 100/1.022 = 97.84 kW.

11. A car engine operates with a thermal efficiency of 35%. Assume the air-conditioner has a coefficient of performance of 3 working as a refrigerator cooling the inside using engine shaft work to drive it. How much fuel energy should be spend extra to remove 1 kJ from the inside?
a) 0.752 kJ
b) 0.952 kJ
c) 0.852 kJ
d) none of the mentioned

Explanation: W = thermal efficiency * Q(fuel) thus Q(fuel) = 1/(0.35*3) = 0.952 kJ.

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