1. A condenser of a refrigeration system rejects heat at a rate of 120kW, while its compressor consumes a power of 30 kW. The coefficient of performance of the system would be:
(a) 1/4
(b) 4
(c) 1/3
(d) 3
(1 Mark, 1992)

Ans: d
Explanation:
The evaporator extracts heat = Heat rejection from the condensor – Power consumed by compressor
=> Q_{e }= 120 – 30 = 90 kW
COP = \frac { { Q }_{ e } }{ W } =\frac { 90 }{ 30 } = 3
2. A reversible heat transfer demands:
(a) The temperature difference causing heat transfer tends to zero.
(b) The system receiving heat must be at a constant temperature.
(c) The system transferring out heat must be at a constant temperature.
(d) Both interacting systems must be at constant temperatures.
(1 Mark, 1993)

Ans: a
3. Any thermodynamic cycle operating between two temperature limits is reversible if the product of efficiency when operating as a heat engine and the coefficient of performance when operating as a refrigeration is equal to 1. (True/ False)
(1 Mark, 1994)

Ans: False
4. Consider a refrigerator and a heat pump working on the reversed Carnot cycle between the same temperature limits. Which of the following is correct?
(a) COP refrigerator = COP of heat pump
(b) COP refrigerator = COP of heat pump + 1
(c) COP refrigerator = COP of heat pump – 1
(d) COP refrigerator = inverse COP of heat pump
(1 Mark, 1995)

Ans: c
5. In the case of a refrigeration system undergoing an irreversible cycle, \oint { \frac { \delta Q }{ T } } is __. (<0/ =0/ >0).
(1 Mark, 1995)

Ans: <0
6. A solar energy based heat engine which receives 80 kJ of heat at 100^{0}C and rejects 70 kJ of heat to the ambient at 30^{0}C is to be designed. The thermal efficiency of the heat engine is
(a) 70%
(b) 1.88%
(c) 12.5%
(d) indeterminate
(2 Mark, 1996)

Ans: c
Explanation:
W = Q_{1 }– Q_{2 }= 80 – 70 = 10 kJ
Efficiency = \frac { W }{ { Q }_{ 1 } } =\frac { 10 }{ 80 } =12.5%
7. A cyclic heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%, the heat rejected per cycle is
(a) 16\frac { 2 }{ 3 } kJ
(b) 33\frac { 1 }{ 3 } kJ
(c) 37\frac { 1 }{ 2 } kJ
(d) 66\frac { 2 }{ 3 } kJ
(2 Mark, 2001)

Ans: a
Explanation:
Efficiency = \frac { Work\quad done }{ Heat\quad supplied }
=> 75% = \frac { 50 }{ { Q }_{ 1 } }
=> Q_{1 }= \frac { 200 }{ 3 } kJ
Heat rejected = Q_{1 }– W = \frac { 200 }{ 3 } 50=16\frac { 2 }{ 3 } kJ
8.A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature reservoir is 727^{0}C, what is the temperature of low temperature reservoir?
(a) 23^{0}C
(b) 23^{0}C
(c) 0^{0}C
(d) 250^{0}C
(2 Mark, 2002)

Ans: b
Explanation:
Carnot efficiency = \frac { { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } }
=> 0.75 = \frac { 1000{ T }_{ 2 } }{ 1000 }
=> T_{2 }= 250 K = – 23^{0}C
9. A solar collector receiving solar radiation at the rate of 0.6 kW/m^{2 }transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 313 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be
(a) 8.33 m^{2}
(b) 16.66 m^{2}
(c) 39.68 m^{2}
(d) 79.36 m^{2}
(2 Mark, 2004)

Ans: d
Explanation:
Let the minimum area of solar collector = A
Given, efficiency = 50%
Heat source (Q_{1}) = (0.6\times 0.5\times A) kW
Now, \frac { { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } } =\frac { W }{ { Q }_{ 1 } }
=> \frac { 350313 }{ 350 } =\frac { 2.5 }{ 0.6\times 0.5\times A }
=> A = 79.36 m^{2}
10. The following four figures have been drawn to represent a fictitious thermodynamics cycle, on the pv and Ts planes.
According to the first law of thermodynamics, equal areas are enclosed by
(a) Figures 1 and 2
(b) Figures 1 and 3
(c) Figures 1 and 4
(d) Figures 2 and 3
(1 Mark, 2005)

Ans: a
Explanation:
We know:
Area under pv diagram represents work transfer.
Area under Ts diagram represents Heat transfer.
According to first law of thermodynamics:
\oint { dQ } =\oint { dW }From the 4 figures given in the question, figure 1 and figure 2 satisfies the first law of thermodynamics because both have same area and same direction (CW).
11. A heat transformer is device that transfers a part of the heat, supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink. In such a heat transformer, 100 kJ of heat is supplied at 350 K. The maximum amount of heat in kJ that can be transferred to 400 K, when the rest is rejected to a heat sink at 300 K is
(a) 12.50
(b) 14.29
(c) 33.33
(d) 57.14
(2 Mark, 2007)
12. An irreversible heat engine extracts heat from a high temperature source at a rate of 100kW and rejects heat to a sink at a rate of 50kW. The entire work output of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high temperature sink is
(a) 50
(b) 250
(c) 300
(d) 360
(2 Mark, 2009)

Ans: c
Explanation:
Work output from heat engine:
W = Q_{1} – Q_{2 }= 100 – 50 = 50 kW
COP of Heat Pump = \frac { 348 }{ 348290 } = 6
Again, COP = \frac { Q }{ W } =\frac { Q }{ 50 }
=> Q = 300 kW
13. A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K. It rejects 1 kJ of heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is
(a) 0.8
(b) 1.0
(c) 1.4
(d) 2.0
(2 Mark, 20014[1])

Ans: c
Explanation:
Using clausius inequality for reversible heat engine:
\frac { { Q }_{ 1 } }{ { T }_{ 1 } } +\frac { { Q }_{ 2 } }{ { T }_{ 2 } } \frac { { Q }_{ 3 } }{ { T }_{ 3 } } =0=> \frac { 2 }{ 1000 } +\frac { { Q }_{ 2 } }{ 800 } \frac { 1 }{ 400 } =0
=> { Q }_{ 2 }=0.4kJ
Now balancing heat and work transfer:
{ Q }_{ 1 }+{ Q }_{ 2 }=W+{ Q }_{ 3 }=> 2 + 0.4 = W + 1
=> W = 1.4 kJ
14. A reversed Carnot cycle refrigerator maintains a temperature of −5° C. The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is ______.
(1 Mark, 20014[4])

Ans: 373.13
Explanation:
COP of refrigerator = \frac { { T }_{ 2 } }{ { T }_{ 1 }{ T }_{ 2 } }
=> COP = \frac { 268 }{ 308268 } = 6.7
Again, COP = \frac { { Q }_{ 2 } }{ W }
=> 6.7 = \frac { 2.5\times { 10 }^{ 3 } }{ W }
=> W = 373.13 Watt
15. A Carnot engine (CE1) works between two temperature reservoirs A and B, where T_{A} = 900 K and T_{B} = 500 K. A second Carnot engine (CE2) works between temperature reservoirs B and C, where T_{C} = 300 K. In each cycle of CE1 and CE2, all the heat rejected by CE1 to reservoir B is used by CE2. For one cycle of operation, if the net Q absorbed by CE1 from reservoir A is 150 MJ, the net heat rejected to reservoir C by CE2 (in MJ) is _____.
(1 Mark, 20015[1])

Ans: 50
Explanation:
For carnot engine (CE – 1):
\eta =1\frac { { T }_{ 2 } }{ { T }_{ 1 } } =1\frac { { Q }_{ 2 } }{ { Q }_{ 1 } }=> \frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { { Q }_{ 2 } }{ { Q }_{ 1 } }
=> \frac { 500 }{ 900 } =\frac { { Q }_{ 2 } }{ 150 }
=> { Q }_{ 2 }=83.33MJ
For carnot engine (CE – 2):
\frac { { T }_{ 3 } }{ { T }_{ 2 } } =\frac { { Q }_{ 3 } }{ { Q }_{ 2 } }=> \frac { 300 }{ 500 } =\frac { { Q }_{ 3 } }{ 83..33 }
=> { Q }_{ 3 }=50MJ
16. A reversible cycle receives 40 kJ of heat from one heat source at a temperature of 127^{0}C and 37 kJ from another heat source at 97^{0} The heat rejected (in kJ) to the heat sink at 47^{0}C is ____.
(2 Mark, 20016[2])

Ans: 64
Explanation:
\sum { \frac { Q }{ T } } =0=> \frac { { Q }_{ 1 } }{ { T }_{ 1 } } +\frac { { Q }_{ 2 } }{ { T }_{ 2 } } +\frac { { Q }_{ 3 } }{ { T }_{ 3 } } =0
=> \frac { 40 }{ 400 } +\frac { 37 }{ 370 } \frac { { Q }_{ 3 } }{ 320 } =0
After solving: { Q }_{ 3 }=64kJ