Second Law of Thermodynamics: Heat Engines, Refrigerators, Reversibility and the Carnot Cycle

This chapter belongs to the Engineering Sciences (XE) paper. It answers Section XE4.4 of XE4 Thermodynamics, one of the nine optional sections, any two of which a candidate chooses in the exam: “Limitations of the first law of thermodynamics, concepts of heat engines and heat pumps/refrigerators, thermal efficiency, coefficient of performance; Kelvin-Planck and Clausius statements and their equivalence; Reversible and irreversible processes; Carnot cycle and Carnot principles/theorems; Thermodynamic temperature scale.” It teaches what the first law cannot say, engines, refrigerators and heat pumps with their efficiency and COP, the two statements of the second law and why they are equivalent, reversibility and its causes, the Carnot cycle and the Carnot theorems, and the Kelvin scale. The numericals are Carnot efficiency and COP between stated reservoirs, a reversible engine driving a reversible refrigerator, the minimum power of a heat pump, the fraction of the Carnot limit that an actual machine reaches, and feasibility checks of claimed machines. Temperatures in every formula are absolute.

1. Limits of the first law; engines, refrigerators and heat pumps

The first law says only that energy is conserved: for a cycle ΔU = 0, so the net heat equals the net work. It never says which way a process runs or how much heat can be turned into work. It permits a hot cup to get hotter by drawing heat from the cooler room, a paddle wheel to be turned by the heat of a stirred liquid, and a ship to run by cooling the sea. None of these is ever seen. Heat flows spontaneously only from hot to cold, work converts to heat completely, and the reverse conversion in a cycle is always partial. The second law supplies this direction and the limit.

A thermal reservoir is a body so large that its temperature stays fixed when it gives or takes heat. A heat engine works in a cycle, takes Q_H from a high-temperature reservoir, delivers net work W and rejects Q_L to a low-temperature reservoir. A refrigerator and a heat pump are the same cycle run backwards: work W goes in, Q_L is taken from the cold space and Q_H is rejected to the warm one. Convention for this chapter: Q_H, Q_L and W are magnitudes (positive numbers), and the first law for the cycle gives W = Q_H − Q_L in every case (W is the output of an engine and the input of a refrigerator or heat pump).

Performance measures of cyclic devices (Q_H, Q_L, W are magnitudes)
DevicePurposeMeasureExpressionRange
Heat engineNet workThermal efficiency ηW/Q_H = 1 − Q_L/Q_H0 < η < 1
RefrigeratorRemove Q_L from cold spaceCOP_RQ_L/W = Q_L/(Q_H − Q_L)Can be below or above 1
Heat pumpDeliver Q_H to warm spaceCOP_HPQ_H/W = Q_H/(Q_H − Q_L)Always above 1
⚠️ COP is not an efficiency
For the same machine between the same two temperatures COP_HP = COP_R + 1, because Q_H = Q_L + W. A heat pump with COP 3 delivers 3 kW of heat for 1 kW of work, which is no violation of the first law since 2 kW comes free from the cold side. An electric resistance heater gives only 1 kW of heat per kW of work, the most that direct conversion of work to heat can give.

2. Kelvin-Planck and Clausius statements and their equivalence

  • Kelvin-Planck statement: it is impossible for a device operating in a cycle to receive heat from a single reservoir and produce a net amount of work. Every engine must therefore reject some heat to a colder reservoir, so η < 1. A machine that breaks it would be a perpetual-motion machine of the second kind.
  • Clausius statement: it is impossible to construct a device operating in a cycle whose sole effect is the transfer of heat from a colder body to a hotter body. A refrigerator therefore needs work, and its COP is finite.

Both statements are negative: they are not proved from other laws but rest on the fact that no violation has ever been observed. They look different yet are equivalent: a device that breaks one can be turned into a device that breaks the other.
(i) Suppose a Kelvin-Planck violator takes Q from a reservoir and turns all of it into work W = Q. Let that work drive an ordinary refrigerator that lifts Q_L from the cold reservoir and rejects Q_L + W to the hot one. The pair together moves Q_L from cold to hot with no net work, which breaks the Clausius statement.
(ii) Suppose a Clausius violator moves Q_L from cold to hot unaided. Run an ordinary engine between the same reservoirs, taking Q_H from the hot one and rejecting exactly Q_L to the cold one. The cold reservoir is unchanged, and the pair takes Q_H − Q_L from the hot reservoir alone and converts it entirely into work, which breaks the Kelvin-Planck statement.

🧠 How to use the statements on a claim
Look for a cycle whose only external effect is one of two things: heat taken from one reservoir and fully converted to work, or heat moved from cold to hot with no work. If either appears the claim is impossible, whatever the numbers. A refrigerator that “runs on its own” and a motor “powered by the heat of the ocean” are the two classic offenders.

3. Reversible and irreversible processes

A process is reversible if it can be reversed so that both the system and the surroundings return exactly to their initial states, leaving no trace anywhere. It needs two things: the system passes through equilibrium states (quasi-equilibrium) and there are no dissipative effects. A process with no irreversibility inside the system boundary is internally reversible; if there is none outside either, for example heat moves only across an infinitesimal temperature difference, it is totally reversible. A reversible process is an idealisation that sets a limit: a reversible work-producing device gives the maximum work and a reversible work-absorbing device needs the minimum. Every real process is irreversible.

  • Friction between solids, and viscous friction in fluids, turns work into heat that cannot be turned back.
  • Unrestrained expansion (a gas rushing into a vacuum) and sudden expansion or compression that leaves the gas without a uniform pressure.
  • Heat transfer through a finite temperature difference, which can never be undone without work.
  • Mixing of different substances or of streams in different states, spontaneous chemical reaction, electrical resistance (Joule heating) and inelastic deformation or hysteresis.
⚠️ Quasi-static is not the same as reversible
A gas compressed very slowly in a cylinder with a rubbing piston is quasi-static yet irreversible, because friction dissipates work. Likewise, slow heat flow across a 50 K gap stays irreversible however slowly it happens. Reversible means quasi-equilibrium and no dissipation, so every reversible process is quasi-static but not the other way round.

4. The Carnot cycle and its efficiency and COP

The Carnot cycle is the reversible cycle between two reservoirs and has four processes: (1→2) reversible isothermal expansion at T_H with heat Q_H absorbed; (2→3) reversible adiabatic (isentropic) expansion that cools the fluid from T_H to T_L; (3→4) reversible isothermal compression at T_L with heat Q_L rejected; (4→1) isentropic compression back to T_H. It can use any working fluid, in a piston-cylinder or in steady flow. On a p-v diagram it is bounded by two isotherms and two adiabats; on a T-s diagram it is a rectangle with T_H on top and T_L at the bottom. Run backwards it is the reversed Carnot cycle, the best possible refrigerator or heat pump.

For an ideal gas: Q_H = nRT_H ln(V₂/V₁) and Q_L = nRT_L ln(V₃/V₄). The two adiabats give T_H V₂^(γ−1) = T_L V₃^(γ−1) and T_H V₁^(γ−1) = T_L V₄^(γ−1), so V₂/V₁ = V₃/V₄ and the logarithms cancel: Q_H/Q_L = T_H/T_L. Therefore

  • η_Carnot = 1 − T_L/T_H
  • COP_R,Carnot = T_L/(T_H − T_L)
  • COP_HP,Carnot = T_H/(T_H − T_L)

Worked example. A Carnot engine works between 727 °C and 27 °C. T_H = 1000 K, T_L = 300 K, so η = 1 − 300/1000 = 0.70. Using the Celsius numbers gives 1 − 27/727 = 0.963, the commonest slip. Efficiency reaches 100% only for T_L = 0 K or T_H → ∞, neither attainable.

5. Carnot principles and comparing the limits

  • First principle: the efficiency of an irreversible engine is always less than that of a reversible engine working between the same two reservoirs.
  • Second principle: all reversible engines between the same two reservoirs have the same efficiency, whatever the working fluid or construction.
  • The same holds for refrigerators and heat pumps: an irreversible one has a lower COP than a reversible one between the same reservoirs, and all reversible ones share one COP.

Proof of the first principle by contradiction. Let an irreversible engine I and a reversible engine R both take Q_H from the hot reservoir, and suppose η_I > η_R, so W_I > W_R. Reverse R, which then needs W_R and returns Q_H to the hot reservoir. The hot reservoir is unchanged, and the pair delivers net work W_I − W_R while taking the same amount of heat from the cold reservoir alone. That breaks the Kelvin-Planck statement, so η_I ≤ η_R. Applying the argument to two reversible engines, each reversed in turn, forces equality.

Which change helps more? Since η = 1 − T_L/T_H, ∂η/∂T_H = T_L/T_H² and ∂η/∂T_L = −1/T_H. As T_L < T_H, T_L/T_H² is smaller than 1/T_H, so lowering the sink temperature by 1 K gains more efficiency than raising the source temperature by 1 K. For a Carnot engine at 600 K and 300 K (η = 0.500), raising the source to 650 K gives 0.538 whereas lowering the sink to 250 K gives 0.583. Note that both changes widen T_H − T_L by 50 K, yet the result differs, because η depends on the ratio T_L/T_H and not on the difference. Practical sinks are fixed by the environment, so engines are improved by raising T_H as far as materials permit. For a refrigerator, COP_R = T_L/(T_H − T_L) rises as the temperature lift T_H − T_L shrinks.

6. The thermodynamic (Kelvin) temperature scale

Ordinary thermometers depend on a substance (mercury expansion, a resistance, a gas) and disagree away from their fixed points. The second law offers a scale that depends on none. By the second Carnot principle the ratio Q_H/Q_L of a reversible engine depends only on the two reservoirs, not on the fluid. Kelvin therefore defined temperature by Q_H/Q_L = T_H/T_L, so that heat ratios measure temperature ratios. One fixed point then fixes the size of the unit: historically the triple point of water was assigned 273.16 K. Since the 2019 revision of the SI the kelvin is defined by fixing the Boltzmann constant at k = 1.380649 × 10⁻²³ J/K, and the scale agrees with the earlier one.

The ideal-gas Carnot cycle above gives Q_H/Q_L = T_H/T_L with T measured on a gas thermometer, so the ideal-gas scale and the thermodynamic scale coincide. The zero of the scale is where Q_L = 0, so η = 1; that would be a Kelvin-Planck violation, so absolute zero cannot be reached, which agrees with the third law. Celsius is t = T − 273.15 (many problems use 273, and the stem then says so).

7. Testing a claimed machine against the second law

Method.
(i) Convert temperatures to kelvin.
(ii) Compute the reversible limit for the same reservoirs: η_max = 1 − T_L/T_H for an engine, COP_max = T_L/(T_H − T_L) for a refrigerator, COP_max = T_H/(T_H − T_L) for a heat pump.
(iii) Compare the claimed value with the limit.
(iv) Check energy balance (Q_H = W + Q_L) separately; the first law and the second law are two independent tests.

Verdict of the comparison
Claimed value against limitVerdict
Above the reversible limitImpossible: violates the second law
Equal to the limitPossible only if the machine is fully reversible
Below the limitPossible, and the machine is irreversible

Worked example. An inventor claims an engine that takes 100 kW from a source at 800 K, rejects heat to a sink at 300 K and produces 70 kW. The claimed η = 0.70, while η_max = 1 − 300/800 = 0.625. It is impossible even though 100 = 70 + 30 satisfies the first law. For a machine that works between three temperatures, as when an engine drives a refrigerator, apply the limits in turn: the reversible engine between T_H and T_0 gives W = Q_H(1 − T_0/T_H), and the reversible refrigerator between T_0 and T_L then removes Q_L = W·T_L/(T_0 − T_L).

Key takeaways

  • The first law conserves energy but gives no direction and no limit; the second law gives both. COP_R = Q_L/W, COP_HP = Q_H/W = COP_R + 1, and η = 1 − Q_L/Q_H, all with Q_H, Q_L, W as magnitudes.
  • Kelvin-Planck: no cyclic device converts heat from a single reservoir wholly into work. Clausius: no cyclic device moves heat from cold to hot as its sole effect. Violating either lets you build a violator of the other, so they are equivalent.
  • Irreversibility comes from friction, unrestrained expansion, heat transfer across a finite ΔT, mixing, reaction and resistance. Reversible means quasi-equilibrium and no dissipation; slow alone is not enough.
  • The Carnot cycle has two isotherms and two isentropes (a rectangle on T-s) and gives Q_H/Q_L = T_H/T_L, hence η = 1 − T_L/T_H, COP_R = T_L/(T_H − T_L), COP_HP = T_H/(T_H − T_L), always in kelvin.
  • Carnot principles: irreversible is worse than reversible between the same reservoirs, and all reversible machines between them are equal, independent of the working fluid. Lowering T_L helps an engine more than raising T_H by the same number of kelvin.
  • The Kelvin scale is defined by Q_H/Q_L = T_H/T_L, independent of any substance, and coincides with the ideal-gas scale. To test a claim, convert to kelvin, compute the reversible limit and compare; a value above the limit is impossible even if the first law balances.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following is the Kelvin-Planck statement of the second law?

    1. Heat cannot flow of its own accord from a colder body to a hotter body
    2. The entropy of an isolated system can never decrease in any process
    3. No device operating in a cycle can receive heat from a single reservoir and produce a net amount of work
    4. Energy can be neither created nor destroyed but only converted from one form to another
    Show answer

    Answer: C — No device operating in a cycle can receive heat from a single reservoir and produce a net amount of work

    The Kelvin-Planck statement forbids a cyclic device from taking heat from one reservoir and delivering net work; it is the statement about engines. The statement about heat flowing from cold to hot unaided is the Clausius statement and is the tempting alternative. The entropy statement is a consequence of the second law, and the energy statement is the first law.

  2. A cyclic device is proposed as follows. Which one violates the Clausius statement of the second law, whatever the temperatures involved?

    1. A refrigerator that takes 1 kW from the cold space and, with no power input, rejects 1 kW to the warm room
    2. A refrigerator that takes 3 kW from the cold space with 1 kW of power input and rejects 4 kW to the warm room
    3. A heat pump that delivers 3 kW to a warm room for 1 kW of power input, taking 2 kW from cold outdoor air
    4. An engine that takes 5 kW from a hot source, rejects 3 kW to a cold sink and delivers 2 kW of work
    Show answer

    Answer: A — A refrigerator that takes 1 kW from the cold space and, with no power input, rejects 1 kW to the warm room

    The only effect of the first device is to move heat from the cold space to the warm room without any work, which is exactly what the Clausius statement forbids. The second and third devices need work input and satisfy the first law; the fourth is an ordinary engine that rejects heat to a sink, so the Kelvin-Planck statement is not violated either. The tempting wrong choice is the heat pump, whose COP of 3 sounds too good but is allowed whenever the temperature lift is small enough.

  3. A Carnot engine receives heat from a source at 727 °C and rejects heat to a sink at 27 °C. Take T(K) = t(°C) + 273. Its thermal efficiency is

    1. 96.3%
    2. 30.0%
    3. 62.5%
    4. 70.0%
    Show answer

    Answer: D — 70.0%

    η = 1 − T_L/T_H = 1 − 300/1000 = 0.70, using kelvin. The value 96.3% results from using the Celsius numbers, 1 − 27/727, which is the standard slip. The value 30.0% is the heat-rejection fraction Q_L/Q_H, not the efficiency.

  4. A Carnot engine works between a 600 K source and a 300 K sink. Two modifications are considered separately: P, raise the source temperature by 50 K; Q, lower the sink temperature by 50 K. Which statement is correct?

    1. P and Q give equal efficiencies, because each widens the temperature difference T_H − T_L by 50 K
    2. P gives the higher efficiency, 58.3%, against 53.8% for Q, because the source is the hotter reservoir
    3. Neither changes the efficiency, since a Carnot engine's efficiency is fixed at 50% for a 2:1 temperature ratio
    4. Q gives the higher efficiency, 58.3%, against 53.8% for P, because η depends on the ratio T_L/T_H
    Show answer

    Answer: D — Q gives the higher efficiency, 58.3%, against 53.8% for P, because η depends on the ratio T_L/T_H

    Case P: η = 1 − 300/650 = 0.538. Case Q: η = 1 − 250/600 = 0.583. Equivalently ∂η/∂T_L = −1/T_H is larger in magnitude than ∂η/∂T_H = T_L/T_H² whenever T_L < T_H. The equal-difference argument is the tempting wrong one: both changes give T_H − T_L = 350 K, but η depends on the ratio, not the difference.

  5. Two reversible engines operate between the same pair of reservoirs, one with an ideal gas and the other with steam as the working fluid. Which statement is correct?

    1. The steam engine is more efficient, since steam can carry more energy per kilogram
    2. Both have the same efficiency, 1 − T_L/T_H, as both are reversible
    3. The ideal-gas engine is more efficient, since an ideal gas has no intermolecular forces to dissipate energy
    4. The efficiencies differ by an amount that depends on the specific heats of the two fluids
    Show answer

    Answer: B — Both have the same efficiency, 1 − T_L/T_H, as both are reversible

    The second Carnot principle says that all reversible engines between the same two reservoirs have the same efficiency, independent of the working fluid. Any difference would let one be reversed against the other and violate the Kelvin-Planck statement. The ideal-gas option is tempting because the textbook derivation uses an ideal gas, but the result is general.

  6. Four machines are claimed, each exchanging heat with only two reservoirs, with temperatures in kelvin. Which claim is consistent with the second law? P: an engine between 500 K and 300 K with η = 45%. Q: a heat pump delivering heat at 300 K from a source at 270 K with COP = 12. R: a refrigerator holding 250 K in surroundings at 300 K with COP = 4. S: an engine between 800 K and 400 K with η = 60%.

    1. P only
    2. Q only
    3. R only
    4. S only
    Show answer

    Answer: C — R only

    Compare each claim with its reversible limit. P: 1 − 300/500 = 40%, and 45% exceeds it. Q: COP_HP,max = 300/30 = 10, and 12 exceeds it. R: COP_R,max = 250/50 = 5, and 4 is below it, so it is possible (and irreversible). S: 1 − 400/800 = 50%, and 60% exceeds it. Only the refrigerator claim is allowed.

  7. The thermodynamic (Kelvin) temperature scale is defined using

    1. the ratio of heats Q_H/Q_L exchanged by a reversible engine between two reservoirs
    2. the thermal expansion of mercury between the freezing and boiling points of water at 1 atm
    3. the pressure of a fixed mass of ideal gas at constant volume, extrapolated to zero pressure
    4. the efficiency of any irreversible heat engine, taken at its maximum power output
    Show answer

    Answer: A — the ratio of heats Q_H/Q_L exchanged by a reversible engine between two reservoirs

    Kelvin defined temperature through Q_H/Q_L = T_H/T_L for a reversible engine; by the second Carnot principle this ratio is independent of the working substance, so the scale depends on no material property. The mercury and gas-thermometer scales are substance-based; the gas scale only coincides with the thermodynamic scale in the ideal-gas limit. An irreversible engine's efficiency is not a property of the reservoirs alone.

  8. Which of the following processes is quasi-static but still irreversible?

    1. Frictionless adiabatic compression of an ideal gas carried out infinitely slowly
    2. Isothermal expansion of a gas in contact with a reservoir at a temperature only infinitesimally higher than the gas
    3. Slow compression of a gas by a piston whose seal rubs on the cylinder wall, the work being dissipated as heat
    4. Free expansion of a gas into an evacuated container through a very small opening
    Show answer

    Answer: C — Slow compression of a gas by a piston whose seal rubs on the cylinder wall, the work being dissipated as heat

    With a rubbing seal the gas can pass through equilibrium states (quasi-static) yet work is dissipated by friction, so the process is irreversible. The frictionless slow adiabatic compression and the isothermal expansion across an infinitesimal temperature difference are reversible. Free expansion is not quasi-static at all, because the gas has no uniform pressure during the rush, so it is not the key.

  9. A reversible heat engine with thermal efficiency 0.25 is run backwards between the same two reservoirs as a refrigerator, the cold reservoir being the engine's sink. The COP of this refrigerator is

    1. 4
    2. 0.75
    3. 3
    4. 1.33
    Show answer

    Answer: C — 3

    For the engine W = 0.25 Q_H and Q_L = 0.75 Q_H. Reversed, the same heats flow the other way, so COP_R = Q_L/W = 0.75/0.25 = 3. The value 4 is COP_HP = Q_H/W of the reversed device, which is exactly COP_R + 1. The value 1.33 is 1/0.75, which has no meaning here.

  10. Which of the following processes are irreversible?

    1. Expansion of a gas into a vacuum after a partition between two rigid insulated chambers is removed
    2. Heat conduction between two bodies whose temperatures differ by only a vanishingly small amount
    3. Mixing of two different gases at the same temperature and pressure after a partition is removed
    4. Frictionless, infinitely slow adiabatic compression of an ideal gas in an insulated cylinder
    Show answer

    Answer: A — Expansion of a gas into a vacuum after a partition between two rigid insulated chambers is removed; C — Mixing of two different gases at the same temperature and pressure after a partition is removed

    Free expansion and the mixing of different gases are spontaneous and cannot be undone without leaving a trace on the surroundings, so both are irreversible. Heat transfer across a vanishingly small temperature difference and frictionless, quasi-equilibrium adiabatic compression are the idealised reversible processes, so they are not keyed.

  11. Which of the following statements about the Carnot cycle are correct?

    1. It consists of two reversible isothermal processes and two reversible adiabatic processes
    2. Its efficiency is higher when steam is the working fluid than when air is the working fluid, for the same two reservoirs
    3. Run in reverse as a refrigerator between T_L and T_H, its COP is T_L/(T_H − T_L)
    4. Its efficiency increases if the heat rejection takes place at a higher temperature T_L
    Show answer

    Answer: A — It consists of two reversible isothermal processes and two reversible adiabatic processes; C — Run in reverse as a refrigerator between T_L and T_H, its COP is T_L/(T_H − T_L)

    The cycle is two isotherms and two isentropes, and reversed it has COP_R = T_L/(T_H − T_L). Efficiency does not depend on the working fluid, so the steam claim is false. Because η = 1 − T_L/T_H, a higher rejection temperature lowers the efficiency, so the statement about a higher rejection temperature is also false.

  12. The Kelvin-Planck and Clausius statements are shown to be equivalent. Which of the following are correct?

    1. A Kelvin-Planck violator driving an ordinary refrigerator gives a pair whose sole effect is heat transfer from cold to hot
    2. A Clausius violator combined with an ordinary engine gives a pair that converts heat from a single reservoir wholly into work
    3. The two statements are independent, so a device may violate one while satisfying the other
    4. The equivalence holds only if the working fluid is an ideal gas
    Show answer

    Answer: A — A Kelvin-Planck violator driving an ordinary refrigerator gives a pair whose sole effect is heat transfer from cold to hot; B — A Clausius violator combined with an ordinary engine gives a pair that converts heat from a single reservoir wholly into work

    Both constructions are the standard proofs: a work-from-one-reservoir device can power a refrigerator to move heat uphill with no net work, and an unaided cold-to-hot heat mover plus an ordinary engine leaves only a single-reservoir heat-to-work conversion. Hence violating one means violating the other, which makes the statements equivalent, not independent. The proofs use no property of the working fluid.

  13. A reversible engine works between a source at 900 K and a sink at 300 K and receives 100 kJ of heat per cycle from the source. All of its work drives a reversible refrigerator that removes heat from a cold space at 250 K and rejects heat to surroundings at 300 K. The heat removed from the cold space per cycle is ____ kJ (answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 333.3

    Engine work W = Q_H(1 − T_L/T_H) = 100 × (1 − 300/900) = 66.67 kJ. Refrigerator COP = T_L/(T_H − T_L) = 250/(300 − 250) = 5. Heat removed Q_L = COP × W = 5 × 66.67 = 333.3 kJ. Checking with the energy balance, the refrigerator rejects 333.3 + 66.7 = 400 kJ to the 300 K surroundings, while the engine rejects 33.3 kJ to the same surroundings.

  14. A house is to be kept at 22 °C when the outdoor temperature is 2 °C, and loses heat at 18 kW. Take T(K) = t(°C) + 273. The minimum electrical power that any heat pump supplying this heat can need, working between the house and the outdoor air, is ____ kW (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.22

    The minimum power corresponds to a reversible (Carnot) heat pump. COP_HP,max = T_H/(T_H − T_L) = 295/(295 − 275) = 14.75. W_min = Q_H/COP = 18/14.75 = 1.22 kW. Electric resistance heating would need 18 kW, nearly fifteen times as much.

  15. A reversible refrigerator with a coefficient of performance of 4 keeps a cold space at −13 °C. Take T(K) = t(°C) + 273. The temperature of the surroundings to which it rejects heat is ____ °C (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 52

    COP_R = T_L/(T_H − T_L) = 4 with T_L = 260 K gives T_H − 260 = 65, so T_H = 325 K. In Celsius, 325 − 273 = 52 °C. The common slip is to use −13 and solve in Celsius, which would give a nonsensical value.

  16. A real engine receives 120 kW of heat from a source at 1000 K and rejects 85 kW to a sink at 300 K. The ratio of its thermal efficiency to the efficiency of a Carnot engine working between the same reservoirs is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.42

    Net work W = 120 − 85 = 35 kW, so η = 35/120 = 0.2917. Carnot efficiency = 1 − 300/1000 = 0.70. Ratio = 0.2917/0.70 = 0.4167, i.e. 0.42. This ratio is the second-law efficiency of the engine: it reaches only about 42% of what the reservoirs allow.