Topic 8.911 min read
Refrigerator and Heat Pump
A refrigerator or heat pump uses external work to transfer heat from a colder reservoir to a hotter reservoir; cooling and heating performance are measured by different coefficients of performance.
Choose Your Level
Beginner Level
A refrigerator uses electrical work to move heat from its cold interior into a warmer room. A heat pump uses the same cycle when warming the room is the useful result.
Recommended for Class 11
Refrigerator and Heat Pump
Overview
A refrigerator or heat pump uses external work to transfer heat from a colder reservoir to a hotter reservoir; cooling and heating performance are measured by different coefficients of performance.
Learning Outcomes
- A refrigerator removes heat Q₂ from a low-temperature region at T₂.
- The compressor supplies work W to the refrigerant.
- The device rejects Q₁ = Q₂ + W to the high-temperature surroundings at T₁.
- Cooling COP is CPcooling = Q₂W.
- Heating COP is CPheating = Q₁W.
- For an ideal reversible device, CPcooling = T₂T₁ − T₂ and CPheating = T₁T₁ − T₂.
Intermediate Level
Reversed Heat-Engine Operation
Natural heat flow is from hot to cold. A refrigerator reverses this direction by supplying work to a compressor. Its circulating working substance, called a refrigerant, extracts heat from the cold region and releases a larger quantity of heat to the warmer surroundings.
Refrigerator and Heat-Pump Viewpoints
| Device name | Desired effect | Useful heat quantity |
|---|---|---|
| refrigerator | remove heat from cold space | Q₂ |
| heat pump | deliver heat to warm space | Q₁ |
Coefficient of Performance for Cooling
Cooling COP is the heat removed from the cold reservoir per unit work supplied.
Coefficient of Performance for Heating
Heating COP is the heat delivered to the warm reservoir per unit work supplied.
Meaning of COP
COP is not an efficiency percentage. A cooling COP of 9 means that every joule of compressor work removes 9 J from the cold region. The device then releases 10 J to the warm surroundings because Q₁ = Q₂ + W.
Textbook Example 8.5
A refrigerator is at 6 °C while the room is at 37 °C. For an ideal refrigerator, find the heat delivered to the room for each joule of electricity consumed.
T₂ = 6 + 273 = 279 K
T₁ = 37 + 273 = 310 K
CPcooling = T₂T₁ − T₂ = 279310 − 279 = 9
For W = 1 J, Q₂ = CPcooling × W = 9 J
Q₁ = Q₂ + W = 9 J + 1 J
Final answer: Q₁ = 10 J delivered to the room per joule of electricity
Cooling COP from Heat Transfers
A refrigerator removes 600 J from its cold space while consuming 200 J of work per cycle. Find Q₁, CPcooling and CPheating.
Q₁ = Q₂ + W = 600 + 200 = 800 J
CPcooling = Q₂W = 600200 = 3
CPheating = Q₁W = 800200 = 4
Final answer: Q₁ = 800 J, CPcooling = 3 and CPheating = 4
Common Mistakes
- Use absolute temperatures in kelvin in ideal COP formulas.
- Cooling COP uses Q₂, while heating COP uses Q₁.
- A COP may be greater than 1 because it compares moved heat with work input; it is not thermal efficiency.
- Heat transfer from cold to hot requires external work.
Short-Question Preparation
Q. Define a refrigerator.
A. A device that removes heat from a region at lower temperature than its surroundings by using external work.
Q. How does a heat pump differ from a refrigerator?
A. It is the same type of device, but the useful output is heat Q₁ delivered to the warm region rather than heat Q₂ removed from the cold region.
Q. Write the energy balance.
A. Q₁ = Q₂ + W, or W = Q₁ − Q₂.
Q. Write both coefficients of performance.
A. CPcooling = Q₂W and CPheating = Q₁W.
Sticky Notes
- A refrigerator removes heat Q₂ from a low-temperature region at T₂.
- The compressor supplies work W to the refrigerant.
- The device rejects Q₁ = Q₂ + W to the high-temperature surroundings at T₁.
- Cooling COP is CPcooling = Q₂W.
- Heating COP is CPheating = Q₁W.
- For an ideal reversible device, CPcooling = T₂T₁ − T₂ and CPheating = T₁T₁ − T₂.
60-Second Revision
A reversed heat engine removes Q₂ from T₂, receives work W and rejects Q₁ to T₁. Q₁ = Q₂ + W. Cooling COP = Q₂W = T₂T₁ − T₂; heating COP = Q₁W = T₁T₁ − T₂.
Summary
A refrigerator or heat pump uses external work to transfer heat from a colder reservoir to a hotter reservoir; cooling and heating performance are measured by different coefficients of performance.
Ask-the-Tutor Modes
- Explain the energy flow simply.
- Compare refrigerator and heat pump.
- Derive both COP formulas.
- Solve a COP numerical.
- Test me with an MCQ.
Exam Focus
Exam Focus
Important points
- A refrigerator removes heat Q₂ from a low-temperature region at T₂.
- The compressor supplies work W to the refrigerant.
- The device rejects Q₁ = Q₂ + W to the high-temperature surroundings at T₁.
- Cooling COP is CPcooling = Q₂W.
- Heating COP is CPheating = Q₁W.
- For an ideal reversible device, CPcooling = T₂T₁ − T₂ and CPheating = T₁T₁ − T₂.
Important formulas
- Q₁ = Q₂ + W
- CPcooling = Q₂W
- CPheating = Q₁W
- CPcooling = T₂T₁ − T₂
- CPheating = T₁T₁ − T₂
Common exam traps
- Use absolute temperatures in kelvin in ideal COP formulas.
- Cooling COP uses Q₂, while heating COP uses Q₁.
- A COP may be greater than 1 because it compares moved heat with work input; it is not thermal efficiency.
- Heat transfer from cold to hot requires external work.
Long question structure
Definition → reversed cycle → energy balance → cooling COP → heating COP → kelvin numerical.
Common Mistakes
Common Mistakes
- Mistake
Use absolute temperatures in kelvin in ideal COP formulas.
- Mistake
Cooling COP uses Q₂, while heating COP uses Q₁.
- Mistake
A COP may be greater than 1 because it compares moved heat with work input; it is not thermal efficiency.
- Mistake
Heat transfer from cold to hot requires external work.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
A refrigerator removes heat Q₂ from a low-temperature region at T₂.
The compressor supplies work W to the refrigerant.
The device rejects Q₁ = Q₂ + W to the high-temperature surroundings at T₁.
Cooling COP is CPcooling = Q₂W.
A refrigerator or heat pump uses external work to transfer heat from a colder reservoir to a hotter reservoir; cooling and heating performance are measured by different coefficients of performance.
- Definition
- A refrigerator or heat pump uses external work to transfer heat from a colder reservoir to a hotter reservoir; cooling and heating performance are measured by different coefficients of performance.
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Exam Practice
Important Question
Q1. Define a refrigerator.
Important Question
Q2. How does a heat pump differ from a refrigerator?
Important Question
Q3. Write the energy balance.
Important Question
Q4. Write both coefficients of performance.