---
title: "Heat pump and COP; second law analysis"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:heat-pump-cop-second-law"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/heat-pump-cop-second-law
site: BCT Engineering
---

# Heat pump and COP; second law analysis

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2068 Magh, 6 marks
- Video: [How Do Refrigerators and Heat Pumps Work? | Thermodynamics | (Solved Examples)](https://www.youtube.com/watch?v=sMHhiXn2fFE) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/heat-pump-cop-second-law
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Define heat pump and its COP. Explain how second law of thermodynamics can be applied to analyze the performance of the heat pump.

## Short answer

A heat pump moves heat from cold to hot using work input; COP = QH/W. The second law limits maximum COP to TH/(TH−TL) (Carnot), requires work input per Clausius statement, and leads to second-law efficiency η_II = COP/COP_max. Entropy generation must be non‑negative.

## Answer

#### Heat pump
A heat pump is a cyclic device that transfers heat from a low-temperature source (outside air, ground, water) to a high-temperature space (a house in winter), using external work. It keeps the warm space warm.

```text
   Warm space T_H  <----- Q_H
                           ^
                         [ HP ] <---- W (input)
                           ^
   Cold source T_L  -----> Q_L
```

#### Coefficient of performance
COP_HP = desired output/required input = Q_H/W = Q_H/(Q_H − Q_L)

It is always greater than 1, because Q_H = Q_L + W. It is related to the refrigerator COP by COP_HP = COP_R + 1.

#### Second law analysis
1. **Clausius statement.** Heat cannot flow from cold to hot without work. So a heat pump cannot work without a work input (W > 0), and its COP is finite.
2. **Maximum COP.** The second law says no heat pump can be better than a reversible one working between the same two temperatures. For a reversible (Carnot) heat pump, Q_H/Q_L = T_H/T_L, so

COP_HP,max = T_H/(T_H − T_L)

   and any real heat pump has COP_HP ≤ T_H/(T_H − T_L). The temperatures are absolute.
3. **Minimum work.** For a heating load Q_H, W_min = Q_H (T_H − T_L)/T_H.
4. **Effect of temperatures.** The smaller the temperature lift (T_H − T_L), the higher the COP. A heat pump is more efficient in mild weather.
5. **Second law efficiency.** η_II = COP_actual/COP_max measures how close a real pump is to the ideal one. The gap comes from irreversibilities (friction, finite ΔT in the heat exchangers, throttling).
6. **Entropy.** For the heat pump together with its two reservoirs, the entropy generated per cycle is S_gen = Q_H/T_H − Q_L/T_L ≥ 0, and S_gen = 0 only for the reversible pump.

Example: house at 293 K, outside at 273 K: COP_max = 293/20 = 14.65. A real unit with COP 4 has η_II = 27 %.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the COP of a heat pump defined?
   A. Q_L/W
   B. W/Q_H
   C. Q_L/Q_H
   D. Q_H/W
   Answer: D. A heat pump's desired output is heat delivered to the warm space, so COP = Q_H/W. (Key point: Heat pump COP)

2. How is COP_HP related to COP_R for the same cycle?
   A. They are equal
   B. COP_HP = COP_R + 1
   C. COP_HP = 1/COP_R
   D. COP_HP = COP_R - 1
   Answer: B. Since Q_H = Q_L + W, dividing by W gives COP_HP = COP_R + 1. (Key point: Relation with refrigerator)

3. The second law limits a real heat pump's COP to:
   A. (T_H - T_L)/T_H
   B. T_H/(T_H - T_L) with absolute temperatures
   C. T_L/(T_H - T_L)
   D. Infinity, since it is a work-saving device
   Answer: B. A reversible heat pump has COP = T_H/(T_H - T_L); no real pump can exceed it. (Key point: Maximum COP)

4. How does a smaller temperature lift (T_H - T_L) affect COP?
   A. COP becomes less than 1
   B. COP increases
   C. No effect
   D. COP decreases
   Answer: B. COP_max = T_H/(T_H - T_L) rises as the lift falls, so heat pumps do better in mild weather. (Key point: Effect of temperature lift)

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