---
title: "Heat engine, heat pump, refrigerator and performance"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:heat-engine-pump-refrigerator"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/heat-engine-pump-refrigerator
site: BCT Engineering
---

# Heat engine, heat pump, refrigerator and performance

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

- Asked in: 2075 Baisakh, 6 marks
- Video: [Heat Engines, Refrigerators, & Cycles: Crash Course Engineering #11](https://www.youtube.com/watch?v=iZOXW5aaCZg) — CrashCourse
- 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-engine-pump-refrigerator
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Define heat engine, heat pump and refrigerator. Also define factors used to measure their performance.

## Short answer

A heat engine receives heat from a hot source, produces net work, and rejects the rest; its performance is thermal efficiency η = W/Q_H. A refrigerator removes heat from a cold space using work input, with COP_R = Q_L/W. A heat pump delivers heat to a warm space using work, with COP_HP = Q_H/W.

## Answer

All three are cyclic devices operating between a high-temperature reservoir (T_H) and a low-temperature reservoir (T_L). They differ in purpose and in the direction of energy flow.

#### Heat engine
A device that operates in a cycle, receives heat Q_H from a high-temperature source, converts part of it to net work W, and rejects the rest Q_L to a low-temperature sink. Examples: steam power plant, petrol engine.

```text
   Hot source T_H
        | Q_H
        v
   [ Heat engine ] ---> W (out)
        | Q_L
        v
   Cold sink T_L
```

**Performance: thermal efficiency**
η = W_net/Q_H = 1 − Q_L/Q_H. For a reversible engine, η = 1 − T_L/T_H.

#### Refrigerator
A cyclic device that removes heat Q_L from a low-temperature space (the cold body), using work input W, and rejects Q_H to the surroundings. Its purpose is to keep the cold space cold. Examples: domestic refrigerator, freezer, air conditioner in cooling mode.

**Performance: coefficient of performance**
COP_R = Q_L/W = Q_L/(Q_H − Q_L). For a reversible refrigerator, COP_R = T_L/(T_H − T_L).

#### Heat pump
A cyclic device that supplies heat Q_H to a warm space (e.g. a room in winter) by absorbing Q_L from a colder source (outside air or ground), using work input W. Its purpose is to keep the warm space warm.

**Performance**
COP_HP = Q_H/W = Q_H/(Q_H − Q_L). For a reversible heat pump, COP_HP = T_H/(T_H − T_L).

#### Relation
COP_HP = COP_R + 1

| Device | Purpose | Performance measure |
|---|---|---|
| Heat engine | Produce work | Thermal efficiency η = W/Q_H (always < 1) |
| Refrigerator | Remove Q_L from cold space | COP_R = Q_L/W (can exceed 1) |
| Heat pump | Deliver Q_H to warm space | COP_HP = Q_H/W (always > 1) |

## Check yourself

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

1. A heat engine:
   A. receives Q_H, converts part to net work and rejects Q_L
   B. absorbs Q_L using work input
   C. receives work and rejects all as heat
   D. operates without a cold sink
   Answer: A. It operates in a cycle between a hot source and a cold sink. (Key point: Heat engine)

2. What distinguishes a heat pump from a refrigerator?
   A. The working fluid
   B. The heat pump delivers Q_H to a warm space, the refrigerator removes Q_L from a cold space
   C. The heat pump uses no work
   D. The refrigerator rejects no heat
   Answer: B. Both are the same cycle with different purpose: COP_HP = Q_H/W, COP_R = Q_L/W. (Key point: Refrigerator vs heat pump purpose)

3. For a reversible refrigerator, COP_R is:
   A. T_H/(T_H − T_L)
   B. 1 − T_L/T_H
   C. T_L/(T_H − T_L)
   D. (T_H − T_L)/T_L
   Answer: C. COP_R = Q_L/(Q_H − Q_L) = T_L/(T_H − T_L). (Key point: Reversible COP of refrigerator)

4. The relation between the COPs for the same cycle is:
   A. COP_HP = COP_R − 1
   B. COP_HP = 1/COP_R
   C. COP_HP = COP_R
   D. COP_HP = COP_R + 1
   Answer: D. Q_H/W = (Q_L + W)/W = COP_R + 1. (Key point: Relation between COPs)

## Related in Fundamental of Thermodynamics & Heat Transfer

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