Second Law of Thermodynamics Questions and Answers

Q: For a refrigerator absorbing heat QL from a cold region and rejecting heat QH to a hot region, the coefficient of performance is written as

The coefficient of performance is \(\frac{Q_{L}}{Q_{H} - Q_{L}}\).A refrigerator's useful effect is the heat removed from the cold space, \(Q_{L}\), and the cost is the net work input, which by the first law is \(W = Q_{H} - Q_{L}\). So \(COP_{R} = Q_{L}/(Q_{H} - Q_{L})\).\(Q_{H}/(Q_{H} - Q_{L})\) is the COP of a heat pump, whose useful effect is the heat delivered. \((Q_{H} - Q_{L})/Q_{L}\) inverts the refrigerator COP, and \(Q_{L}/Q_{H}\) does not involve the work at all.

Q: A reversible engine receives 75 kJ/s of energy from a reservoir at 750 K and does 12 kJ/s of work. The heat is rejected to two reservoirs at 650 K and 550 K. The rate of heat rejection (in kJ/s) to the reservoir at 650 K is

The heat rejected to the 650 K reservoir is 52 kJ/s.Energy balance: the total heat rejected is \(75 - 12 = 63\) kJ/s. A reversible engine produces no entropy, so the entropy taken from the hot reservoir equals the entropy given to the two cold ones. With \(Q_{1}\) to 650 K and \(Q_{2} = 63 - Q_{1}\) to 550 K:\[ \begin{aligned} \frac{75}{750} &= \frac{Q_{1}}{650} + \frac{63 - Q_{1}}{550} \\ 0.1 - 0.11455 &= Q_{1}\left(\frac{1}{650} - \frac{1}{550}\right) \\ Q_{1} &= \frac{-0.01455}{-2.797\times10^{-4}} = 52.0 \end{aligned} \]So 52 kJ/s goes to 650 K and 11 kJ/s to 550 K. The option 11 is the heat to the other reservoir; 31 and 41 do not satisfy the entropy balance.

Q: If \(Q_{L}\) represents the magnitude of heat transfer from a low temperature reservoir to a cyclic device and \(Q_{H}\) represents the magnitude of heat transfer from a cyclic device to a high temperature reservoir, then for the same \(Q_{L}\) and \(Q_{H}\), the coefficient performance of a refrigerator \((COP_{R})\) and the coefficient performance of a heat pump \((COP_{HP})\) can be related as

The relation is \(COP_{HP} = COP_{R} + 1\).For the same device, with work input \(W = Q_{H} - Q_{L}\):\[ \begin{aligned} COP_{R} &= \frac{Q_{L}}{Q_{H} - Q_{L}} \\ COP_{HP} &= \frac{Q_{H}}{Q_{H} - Q_{L}} = \frac{Q_{L} + (Q_{H} - Q_{L})}{Q_{H} - Q_{L}} = COP_{R} + 1 \end{aligned} \]A heat pump delivers all the heat extracted from the cold side plus the work input, which is why its COP is always one more than the refrigerator's. The other relations do not follow; in particular the product of the two COPs is not 1, and \(COP_{HP}\) can never be less than \(COP_{R}\).

Q: Consider two Carnot heat engines A and B operating in series. Engine A receives heat from a reservoir at 1750 K and rejects heat to another reservoir at temperature T. Engine B receives an amount of energy same as that rejected by Engine A from the reservoir at temperature T. Engine B then rejects heat to another reservoir at 320 K. In both cases, the engines produce some amount of work. If the thermal efficiencies of both the engines are the same, then the temperature T is approximately

T is approximately 748 K.Carnot efficiencies depend only on the reservoir temperatures. Setting them equal:\[ \begin{aligned} 1 - \frac{T}{1750} &= 1 - \frac{320}{T} \\ T^{2} &= 1750 \times 320 = 560000 \\ T &\approx 748.3\ \text{K} \end{aligned} \]So the intermediate temperature is the geometric mean of the extremes, not their arithmetic mean (which would be 1035 K). The other options are 100 K steps away and give unequal efficiencies: at 848 K, for example, engine A runs at 51.5% and engine B at 62.3%.

Q: The Clausius inequality holds good for

The Clausius inequality holds for any cycle.The inequality states that for a system going through a complete cycle,\[ \oint \frac{\delta Q}{T} \le 0 \]with equality for a reversible cycle and strict inequality for an irreversible one. It is a statement about cycles, reversible or not, and it is what leads to the definition of entropy.It is not stated for an arbitrary process, which need not return to its starting state. Restricting it to reversible processes or reversible cycles throws away the inequality, which matters precisely for irreversible cycles.

Q: A heat engine extracts heat (\(Q_{H}\)) from a thermal reservoir at a temperature of 1000 K and rejects heat (\(Q_{L}\)) to a thermal reservoir at a temperature of 100 K, while producing work (\(W\)). Which one of the combinations of [\(Q_{H}\), \(Q_{L}\) and \(W\)] given is allowed?

The allowed combination is \(Q_{H}\) = 2000 J, \(Q_{L}\) = 750 J, \(W\) = 1250 J.A heat engine must satisfy the first law, \(Q_{H} = Q_{L} + W\), and the second law, which caps its efficiency at the Carnot value:\[ \begin{aligned} \eta_{max} &= 1 - \frac{T_{L}}{T_{H}} \\ &= 1 - \frac{100}{1000} = 0.9 \end{aligned} \]2000, 750, 1250: \(750 + 1250 = 2000\), and \(\eta = 1250/2000 = 0.625 \lt 0.9\). Allowed.2000, 500, 1000: \(500 + 1000 = 1500 \ne 2000\); energy is not conserved.6000, 500, 5500: energy balances, but \(\eta = 5500/6000 \approx 0.917 \gt 0.9\), better than a Carnot engine, which the second law forbids.6000, 600, 5500: \(600 + 5500 = 6100 \ne 6000\); more energy leaves than enters.

Q: A furnace can supply heat steadily at 1200 K at a rate of 24000 kJ/min. The maximum amount of power (in kW) that can be produced by using the heat supplied by this furnace in an environment at 300 K is

The maximum power is 300 kW.The most work any engine can extract from heat at 1200 K, rejecting to surroundings at 300 K, is that of a reversible (Carnot) engine:\[ \begin{aligned} \eta_{max} &= 1 - \frac{T_{L}}{T_{H}} = 1 - \frac{300}{1200} = 0.75 \\ \dot{Q} &= \frac{24000}{60} = 400\ \text{kW} \\ \dot{W}_{max} &= 0.75 \times 400 = 300\ \text{kW} \end{aligned} \]18000 is 0.75 times the heat rate left in kJ/min, without converting to kilowatts. Zero would apply only if the furnace and the environment were at the same temperature, and 150 kW is half the Carnot limit with no basis in the data.

Q: For the isothermal expansion of an ideal gas in a piston cylinder system, the net heat supplied during the process is equal to the net work. Which one of the following statements is correct for the given process?

The process is possible.For an ideal gas the internal energy depends only on temperature, so in an isothermal process \(\Delta U = 0\). The first law then gives \(Q = W\): all the heat supplied is converted into expansion work. That is exactly what the question describes.This does not violate the second law, which forbids a cycle converting heat completely into work; a single non-cyclic process such as this expansion can do so, because the gas ends in a different state. So the options claiming a violation of either law are wrong.

Q: The temperature of the working fluid in a real heat engine cycle changes during heat addition and heat rejection processes. The maximum and minimum temperatures of the cycle are \(T_{\max}\) and \(T_{\min}\), respectively. If \(\eta_{C}\) is the thermal efficiency of a Carnot engine operating between these temperature limits, then the thermal efficiency, \(\eta\) of the real heat engine satisfies the relation

The real engine satisfies \(\eta \lt \eta_{C}\).The Carnot efficiency \(\eta_{C} = 1 - T_{min}/T_{max}\) is the highest possible for any engine working between these temperature limits. A real engine falls short for two reasons: it is irreversible, and its heat is added and rejected over a range of temperatures, so the mean temperature of heat addition is below \(T_{max}\) and the mean temperature of rejection is above \(T_{min}\).\(\eta \gt \eta_{C}\) would violate the second law, \(\eta = \eta_{C}\) needs a fully reversible engine with heat exchange only at the extremes, and \(1 + \eta_{C}\) exceeds 100%.

Direction for [ Question No: 1 To 2 ] :

Notation used:
P : pressure, V : volume, T : temperature, s : specific entropy, h : specific enthalpy, C_p : molar heat capacity at constant pressure, C_v : molar heat capacity at constant volume, R : universal gas constant

For a refrigerator absorbing heat Q_L from a cold region and rejecting heat Q_H to a hot region, the coefficient of performance is written as

A.\(\frac{Q_{L}}{Q_{H} - Q_{L}}\)
B.\(\frac{Q_{H}}{Q_{H} - Q_{L}}\)
C.\(\frac{Q_{H} - Q_{L}}{Q_{L}}\)
D.\(\frac{Q_{L}}{Q_{H}}\)

Exam Year(s)

GATE XE - 2018,

Answer: A

The coefficient of performance is \(\frac{Q_{L}}{Q_{H} - Q_{L}}\).

A refrigerator's useful effect is the heat removed from the cold space, \(Q_{L}\), and the cost is the net work input, which by the first law is \(W = Q_{H} - Q_{L}\). So \(COP_{R} = Q_{L}/(Q_{H} - Q_{L})\).

\(Q_{H}/(Q_{H} - Q_{L})\) is the COP of a heat pump, whose useful effect is the heat delivered. \((Q_{H} - Q_{L})/Q_{L}\) inverts the refrigerator COP, and \(Q_{L}/Q_{H}\) does not involve the work at all.

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A reversible engine receives 75 kJ/s of energy from a reservoir at 750 K and does 12 kJ/s of work. The heat is rejected to two reservoirs at 650 K and 550 K. The rate of heat rejection (in kJ/s) to the reservoir at 650 K is

A.11
B.31
C.41
D.52

Exam Year(s)

GATE XE - 2018,

Answer: D

The heat rejected to the 650 K reservoir is 52 kJ/s.

Energy balance: the total heat rejected is \(75 - 12 = 63\) kJ/s. A reversible engine produces no entropy, so the entropy taken from the hot reservoir equals the entropy given to the two cold ones. With \(Q_{1}\) to 650 K and \(Q_{2} = 63 - Q_{1}\) to 550 K:

\[ \begin{aligned} \frac{75}{750} &= \frac{Q_{1}}{650} + \frac{63 - Q_{1}}{550} \\ 0.1 - 0.11455 &= Q_{1}\left(\frac{1}{650} - \frac{1}{550}\right) \\ Q_{1} &= \frac{-0.01455}{-2.797\times10^{-4}} = 52.0 \end{aligned} \]

So 52 kJ/s goes to 650 K and 11 kJ/s to 550 K. The option 11 is the heat to the other reservoir; 31 and 41 do not satisfy the entropy balance.

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Direction for [ Question No: 3 To 4 ] :

Note: Usual notations have been used for thermodynamic variables.

Useful Data: Unless otherwise specified, the following data may be assumed. Universal gas constant, \(\bar{R}\) = 8.314 kJ/kmol.K; Acceleration due to gravity, g = 9.81 m/s²; Molecular mass of air, M_air = 29 kg/kmol; Specific heat of air at constant pressure, c_p = 1.005 kJ/kg.K; Ratio of specific heats of air, γ = 1.4. Assume air to be a perfect gas unless specified otherwise.

If \(Q_{L}\) represents the magnitude of heat transfer from a low temperature reservoir to a cyclic device and \(Q_{H}\) represents the magnitude of heat transfer from a cyclic device to a high temperature reservoir, then for the same \(Q_{L}\) and \(Q_{H}\), the coefficient performance of a refrigerator \((COP_{R})\) and the coefficient performance of a heat pump \((COP_{HP})\) can be related as

A.\(COP_{R} = 1 - COP_{HP}\)
B.\(COP_{HP} = COP_{R} + 1\)
C.\(COP_{R} \cdot COP_{HP} = 1\)
D.\(COP_{HP} = COP_{R} - 1\)

Exam Year(s)

GATE XE - 2012,

Answer: B

The relation is \(COP_{HP} = COP_{R} + 1\).

For the same device, with work input \(W = Q_{H} - Q_{L}\):

\[ \begin{aligned} COP_{R} &= \frac{Q_{L}}{Q_{H} - Q_{L}} \\ COP_{HP} &= \frac{Q_{H}}{Q_{H} - Q_{L}} = \frac{Q_{L} + (Q_{H} - Q_{L})}{Q_{H} - Q_{L}} = COP_{R} + 1 \end{aligned} \]

A heat pump delivers all the heat extracted from the cold side plus the work input, which is why its COP is always one more than the refrigerator's. The other relations do not follow; in particular the product of the two COPs is not 1, and \(COP_{HP}\) can never be less than \(COP_{R}\).

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Consider two Carnot heat engines A and B operating in series. Engine A receives heat from a reservoir at 1750 K and rejects heat to another reservoir at temperature T. Engine B receives an amount of energy same as that rejected by Engine A from the reservoir at temperature T. Engine B then rejects heat to another reservoir at 320 K. In both cases, the engines produce some amount of work. If the thermal efficiencies of both the engines are the same, then the temperature T is approximately

A.848 K
B.748 K
C.648 K
D.548 K

Exam Year(s)

GATE XE - 2012,

Answer: B

T is approximately 748 K.

Carnot efficiencies depend only on the reservoir temperatures. Setting them equal:

\[ \begin{aligned} 1 - \frac{T}{1750} &= 1 - \frac{320}{T} \\ T^{2} &= 1750 \times 320 = 560000 \\ T &\approx 748.3\ \text{K} \end{aligned} \]

So the intermediate temperature is the geometric mean of the extremes, not their arithmetic mean (which would be 1035 K). The other options are 100 K steps away and give unequal efficiencies: at 848 K, for example, engine A runs at 51.5% and engine B at 62.3%.

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Which equation states that the process involving the interaction of a system and the surroundings takes place only if the net entropy of the combined system increases or in the limit remains constant?

A.\((dS)_{isolated} \geq 0\)
B.\((dS)_{system} \geq 0\)
C.\((dS)_{surroundings} \geq 0\)
D.\((dS)_{universe} \geq 0\)

Exam Year(s)

DSSSB - 2023,

Answer: D

The equation is \((dS)_{universe} \geq 0\). The principle of increase of entropy says that a process involving a system and its surroundings can take place only if the net entropy of the system and surroundings together (the universe) increases, or remains constant in the limit of a reversible process. The entropy of the system or of the surroundings taken alone may fall, so the inequalities of those quantities need not hold. The isolated system statement is also true in general but the question speaks of the system and the surroundings together, so the universe is the keyed quantity.

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The Clausius inequality holds good for

A.any process
B.any cycle
C.only reversible process
D.only reversible cycle

Exam Year(s)

GATE ME - 2022,

Answer: B

The Clausius inequality holds for any cycle.

The inequality states that for a system going through a complete cycle,

\[ \oint \frac{\delta Q}{T} \le 0 \]

with equality for a reversible cycle and strict inequality for an irreversible one. It is a statement about cycles, reversible or not, and it is what leads to the definition of entropy.

It is not stated for an arbitrary process, which need not return to its starting state. Restricting it to reversible processes or reversible cycles throws away the inequality, which matters precisely for irreversible cycles.

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A heat engine extracts heat (\(Q_{H}\)) from a thermal reservoir at a temperature of 1000 K and rejects heat (\(Q_{L}\)) to a thermal reservoir at a temperature of 100 K, while producing work (\(W\)). Which one of the combinations of [\(Q_{H}\), \(Q_{L}\) and \(W\)] given is allowed?

A.\(Q_{H}\) = 2000 J, \(Q_{L}\) = 500 J, \(W\) = 1000 J
B.\(Q_{H}\) = 2000 J, \(Q_{L}\) = 750 J, \(W\) = 1250 J
C.\(Q_{H}\) = 6000 J, \(Q_{L}\) = 500 J, \(W\) = 5500 J
D.\(Q_{H}\) = 6000 J, \(Q_{L}\) = 600 J, \(W\) = 5500 J

Exam Year(s)

GATE ME - 2023,

Answer: B

The allowed combination is \(Q_{H}\) = 2000 J, \(Q_{L}\) = 750 J, \(W\) = 1250 J.

A heat engine must satisfy the first law, \(Q_{H} = Q_{L} + W\), and the second law, which caps its efficiency at the Carnot value:

\[ \begin{aligned} \eta_{max} &= 1 - \frac{T_{L}}{T_{H}} \\ &= 1 - \frac{100}{1000} = 0.9 \end{aligned} \]

2000, 750, 1250: \(750 + 1250 = 2000\), and \(\eta = 1250/2000 = 0.625 \lt 0.9\). Allowed.
2000, 500, 1000: \(500 + 1000 = 1500 \ne 2000\); energy is not conserved.
6000, 500, 5500: energy balances, but \(\eta = 5500/6000 \approx 0.917 \gt 0.9\), better than a Carnot engine, which the second law forbids.
6000, 600, 5500: \(600 + 5500 = 6100 \ne 6000\); more energy leaves than enters.

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A furnace can supply heat steadily at 1200 K at a rate of 24000 kJ/min. The maximum amount of power (in kW) that can be produced by using the heat supplied by this furnace in an environment at 300 K is

A.300
B.150
C.18000
D.0

Exam Year(s)

GATE ME - 2024,

Answer: A

The maximum power is 300 kW.

The most work any engine can extract from heat at 1200 K, rejecting to surroundings at 300 K, is that of a reversible (Carnot) engine:

\[ \begin{aligned} \eta_{max} &= 1 - \frac{T_{L}}{T_{H}} = 1 - \frac{300}{1200} = 0.75 \\ \dot{Q} &= \frac{24000}{60} = 400\ \text{kW} \\ \dot{W}_{max} &= 0.75 \times 400 = 300\ \text{kW} \end{aligned} \]

18000 is 0.75 times the heat rate left in kJ/min, without converting to kilowatts. Zero would apply only if the furnace and the environment were at the same temperature, and 150 kW is half the Carnot limit with no basis in the data.

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For the isothermal expansion of an ideal gas in a piston cylinder system, the net heat supplied during the process is equal to the net work. Which one of the following statements is correct for the given process?

A.The process is possible.
B.The process is not possible as it violates the second law of thermodynamics.
C.The process is not possible as it violates the first law of thermodynamics.
D.The process is possible, however, it violates the second law of thermodynamics.

Exam Year(s)

GATE ME - 2026,

Answer: A

The process is possible.

For an ideal gas the internal energy depends only on temperature, so in an isothermal process \(\Delta U = 0\). The first law then gives \(Q = W\): all the heat supplied is converted into expansion work. That is exactly what the question describes.

This does not violate the second law, which forbids a cycle converting heat completely into work; a single non-cyclic process such as this expansion can do so, because the gas ends in a different state. So the options claiming a violation of either law are wrong.

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10.

The temperature of the working fluid in a real heat engine cycle changes during heat addition and heat rejection processes. The maximum and minimum temperatures of the cycle are \(T_{\max}\) and \(T_{\min}\), respectively. If \(\eta_{C}\) is the thermal efficiency of a Carnot engine operating between these temperature limits, then the thermal efficiency, \(\eta\) of the real heat engine satisfies the relation

A.\(\eta \gt \eta_{C}\)
B.\(\eta \lt \eta_{C}\)
C.\(\eta = \eta_{C}\)
D.\(\eta = 1 + \eta_{C}\)

Exam Year(s)

GATE XE - 2022,

Answer: B

The real engine satisfies \(\eta \lt \eta_{C}\).

The Carnot efficiency \(\eta_{C} = 1 - T_{min}/T_{max}\) is the highest possible for any engine working between these temperature limits. A real engine falls short for two reasons: it is irreversible, and its heat is added and rejected over a range of temperatures, so the mean temperature of heat addition is below \(T_{max}\) and the mean temperature of rejection is above \(T_{min}\).

\(\eta \gt \eta_{C}\) would violate the second law, \(\eta = \eta_{C}\) needs a fully reversible engine with heat exchange only at the extremes, and \(1 + \eta_{C}\) exceeds 100%.

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Second Law of Thermodynamics Questions and Answers

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