Refrigerators and Heat Pumps

Refrigerators and heat pumps are devices that use external work to transfer thermal energy from a colder region to a warmer region.

This is opposite to the natural direction of spontaneous heat flow:

hot → cold

To force energy to move:

cold → hot

the device must receive energy as work, usually from electricity.

A refrigerator and a heat pump operate using essentially the same thermodynamic principle. The main difference is which energy transfer we consider useful.

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Natural Heat Flow

If a hot object and a cold object are placed in thermal contact, energy naturally transfers:

hot → cold

This process is spontaneous and consistent with the:

Second Law of Thermodynamics

A refrigerator must do the opposite.

It removes thermal energy from a colder space and transfers it into warmer surroundings.

Therefore:

external work is required

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Basic Refrigerator Energy Flow

A refrigerator operates between:

• a cold reservoir
• a hot reservoir

The refrigerator removes thermal energy:

Q_c

from the cold region.

Work:

W

is supplied to the refrigerator.

The refrigerator then releases:

Q_h

to the warmer surroundings.

Energy conservation gives:

Q_h = Q_c + W

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Energy Is Conserved

Suppose a refrigerator removes:

500 J

from its interior and receives:

200 J

of electrical work.

Then:

Q_h = Q_c + W

Q_h = 500 + 200

Q_h = 700 J

Therefore the refrigerator releases:

700 J

into the room.

This is more energy than it removed from inside because the electrical work also ultimately contributes energy.

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Why the Back of a Refrigerator Feels Warm

A refrigerator does not simply "destroy" thermal energy inside the cabinet.

It transfers that energy into the room.

The coils at the back or underneath the refrigerator release:

Q_h

which includes:

energy removed from food + electrical work supplied

Therefore the condenser coils can feel warm or hot.

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Why Work Is Required

The spontaneous direction of heat transfer is:

hot → cold

A refrigerator forces energy to move:

cold → hot

This cannot happen continuously by itself.

External work must therefore be supplied.

Usually:

electrical energy → compressor work

The compressor is one of the most important components of a refrigerator or heat pump.

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The Refrigeration Cycle

A typical vapor-compression refrigerator contains four major components:

• evaporator
• compressor
• condenser
• expansion valve

A refrigerant circulates continuously through these components.

The basic cycle is:

evaporator → compressor → condenser → expansion valve → evaporator

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The Refrigerant

A refrigerant is a fluid that circulates through the system.

It is chosen because it can:

• evaporate at relatively low temperatures
• condense at suitable higher temperatures
• transport substantial thermal energy
• operate through repeated pressure changes

The refrigerant repeatedly changes:

pressure

temperature

and often:

state of matter

as it moves through the cycle.

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Stage 1: Evaporator

The refrigerant enters the evaporator at relatively:

low pressure and low temperature

It absorbs thermal energy from the refrigerated space.

Therefore:

cold space → refrigerant

As it absorbs energy, the refrigerant evaporates.

The energy absorbed is:

Q_c

This is the useful effect of a refrigerator.

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Why Evaporation Is Useful

Evaporation requires energy.

During the phase change:

liquid → gas

the refrigerant absorbs latent heat.

This allows the refrigerant to remove substantial amounts of thermal energy without requiring an enormous temperature increase.

The evaporator therefore cools the refrigerator interior.

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Stage 2: Compressor

The refrigerant gas then enters the:

compressor

The compressor uses electrical energy to perform work on the gas.

Compression causes:

• pressure to increase
• temperature to increase

The refrigerant leaves the compressor as a:

hot, high-pressure gas

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Why Compression Raises Temperature

When work is done on a gas, its internal energy can increase.

Therefore its temperature rises.

The compressor raises the refrigerant temperature above the temperature of the surrounding room.

This is essential because thermal energy must eventually flow:

refrigerant → room

For that to happen spontaneously, the refrigerant must be hotter than its surroundings.

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Stage 3: Condenser

The hot, high-pressure refrigerant enters the:

condenser

The condenser is usually located where it can transfer energy to the surrounding air.

Because:

T_refrigerant > T_room

thermal energy flows:

refrigerant → room

The refrigerant cools and condenses:

gas → liquid

The energy released is:

Q_h

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Condensation and Latent Heat

During condensation, the refrigerant releases latent heat.

Therefore the condenser can transfer substantial thermal energy to the surroundings.

This explains why the outside coils of a refrigeration system become warm.

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Stage 4: Expansion Valve

The high-pressure liquid refrigerant then passes through an:

expansion valve

or similar restriction.

Its pressure drops dramatically.

This lowers its temperature.

The refrigerant becomes a cold, low-pressure mixture ready to return to the evaporator.

The cycle then repeats.

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The Complete Refrigeration Cycle

The four main stages are:

Evaporator

absorbs Q_c from cold space

Compressor

work W is supplied

Condenser

releases Q_h to warm surroundings

Expansion valve

reduces pressure and temperature

Then:

cycle repeats

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Refrigerator Energy Diagram

The overall energy flow is:

cold reservoir

↓

Q_c

↓

refrigerator

← W

↓

Q_h

↓

hot reservoir

Energy conservation requires:

Q_h = Q_c + W

---

Refrigerators and the Second Law

At first, refrigerators might seem to violate the Second Law because they move energy:

cold → hot

But the Second Law does not say this is impossible.

It says such a transfer cannot occur as the sole result without other changes.

A refrigerator requires:

work input

and the complete process increases total entropy.

Therefore refrigerators are completely consistent with the Second Law.

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Refrigerator vs Heat Engine

A heat engine and refrigerator operate in opposite thermodynamic directions.

Heat engine

hot → cold

while producing:

work output

Refrigerator

Uses:

work input

to transfer energy:

cold → hot

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Comparison

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Heat Pumps

A heat pump operates on essentially the same physical cycle as a refrigerator.

The major difference is:

which energy transfer is useful

For a refrigerator, the desired result is:

removing Q_c from the cold region

For a heat pump, the desired result is:

delivering Q_h to the warm region

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Refrigerator

The goal is to keep something:

cold

Examples include:

• refrigerator
• freezer
• cold-storage room
• food display cooler

The useful energy transfer is:

Q_c

---

Heat Pump

The goal is to keep something:

warm

For example, a heat pump may remove thermal energy from cold outdoor air and transfer it into a building.

The useful energy transfer is:

Q_h

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Same Machine, Different Purpose

Imagine a device between:

cold outdoors

and:

warm house

It extracts Q_c from outdoors.

It receives electrical work W.

It delivers:

Q_h = Q_c + W

into the house.

When the desired effect is heating the house, we call the device a:

heat pump

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Air Conditioners

An air conditioner is thermodynamically very similar to a refrigerator.

Its useful purpose is:

removing thermal energy from a building

Therefore it operates like a large refrigerator.

Indoor air acts as the region being cooled.

Outdoor air receives the rejected thermal energy.

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Reversible Heat Pumps

Many modern systems can reverse their operating direction.

During summer:

inside → outside

The system cools the building.

During winter:

outside → inside

The system heats the building.

This is commonly called a:

reversible heat pump

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How Can a Heat Pump Get Energy from Cold Air?

A common misconception is that cold air contains no thermal energy.

Even cold outdoor air has internal energy as long as its temperature is above absolute zero.

A heat pump can use a refrigerant that is even colder than the outdoor air.

Therefore thermal energy can spontaneously transfer:

outdoor air → colder refrigerant

The compressor then raises the refrigerant's temperature so energy can be released indoors.

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Example: Heating a House

Suppose:

outside air = 5°C

The refrigerant in the outdoor evaporator might be colder than 5°C.

Therefore:

outside air → refrigerant

Thermal energy enters the refrigerant.

The compressor then raises its temperature.

If the refrigerant reaches a temperature above the indoor temperature:

refrigerant → indoor air

Energy can then be delivered into the house.

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Coefficient of Performance

Refrigerators and heat pumps are not normally described using heat-engine efficiency.

Instead, we use the:

coefficient of performance

abbreviated:

COP

COP compares the desired thermal-energy transfer with the work required.

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Refrigerator COP

For a refrigerator, the desired effect is removing energy from the cold region.

Therefore:

COP_R = Q_c/W

where:

• COP_R = refrigerator coefficient of performance
• Q_c = thermal energy removed from cold region
• W = work input

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Worked Example: Refrigerator COP

A refrigerator removes:

Q_c = 900 J

while requiring:

W = 300 J

Then:

COP_R = 900/300

COP_R = 3.0

This means the refrigerator removes:

3 J of thermal energy

from its cold interior for every:

1 J of work supplied

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COP Can Be Greater Than 1

This is very important.

Suppose:

COP = 4

This does not mean:

400% efficiency

COP is not the same as efficiency.

The device is not converting 1 J into 4 J.

Instead, 1 J of work allows it to:

transfer additional energy from another location

Energy remains conserved.

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Heat Pump COP

For a heat pump, the desired effect is delivering thermal energy to the warm region.

Therefore:

COP_HP = Q_h/W

Since:

Q_h = Q_c + W

a heat pump can deliver more thermal energy than the electrical work it consumes.

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Worked Example: Heat Pump COP

A heat pump receives:

W = 1000 J

and extracts:

Q_c = 3000 J

from outdoors.

Then:

Q_h = Q_c + W

Q_h = 3000 + 1000

Q_h = 4000 J

Therefore:

COP_HP = 4000/1000

COP_HP = 4.0

For every 1 J of electrical work supplied, the system delivers 4 J of thermal energy indoors.

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Relationship Between the COPs

Because:

Q_h = Q_c + W

divide everything by W:

Q_h/W = Q_c/W + 1

Therefore:

COP_HP = COP_R + 1

for the same idealized device operating between the same two reservoirs.

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Worked Example

Suppose a refrigeration system has:

COP_R = 3.5

Then:

COP_HP = COP_R + 1

COP_HP = 3.5 + 1

COP_HP = 4.5

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COP and Temperature Difference

A refrigerator or heat pump generally performs better when the temperature difference between hot and cold regions is:

smaller

Why?

Because moving thermal energy across a smaller temperature difference requires less work.

Therefore:

smaller temperature difference → higher possible COP

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Ideal Refrigerator COP

For an ideal reversible refrigerator operating between T_c and T_h:

COP_R,max = T_c/(T_h − T_c)

Temperatures must be in:

kelvin

This gives the maximum theoretical refrigerator COP.

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Ideal Heat Pump COP

For an ideal reversible heat pump:

COP_HP,max = T_h/(T_h − T_c)

Again:

temperatures must be in kelvin

These are theoretical maximum values based on reversible operation.

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Worked Example: Ideal Refrigerator

Suppose:

T_c = 270 K

T_h = 300 K

Then:

COP_R,max = 270/(300 − 270)

COP_R,max = 270/30

COP_R,max = 9.0

This is the ideal maximum.

A real refrigerator would have a lower COP.

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Worked Example: Ideal Heat Pump

Using the same temperatures:

COP_HP,max = 300/(300 − 270)

COP_HP,max = 300/30

COP_HP,max = 10.0

Notice:

COP_HP = COP_R + 1

10 = 9 + 1

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Larger Temperature Difference

Suppose:

T_c = 250 K

T_h = 300 K

Then:

COP_R,max = 250/(300 − 250)

COP_R,max = 250/50

COP_R,max = 5.0

The larger temperature difference reduced the maximum COP.

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Heat Pumps in Cold Weather

As outdoor temperature falls, a heat pump must move thermal energy across a larger temperature difference.

Therefore its COP generally:

decreases

This is one reason heat pumps can require more electrical energy during very cold weather.

System design and heat-pump type also strongly affect real performance.

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Why Heat Pumps Can Be Efficient for Heating

Consider ordinary electric resistance heating.

Approximately:

1 J electrical energy → 1 J thermal energy

at the point of use.

A heat pump does something different.

It uses electrical work to move additional thermal energy from outdoors.

For example:

1 J electrical work

•  

3 J transferred from outdoors

=

4 J delivered indoors

Therefore:

COP_HP = 4

This can make heat pumps highly effective for space heating.

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Household Refrigerator

A household refrigerator uses the vapor-compression cycle to maintain a low interior temperature.

The main components include:

• compressor
• condenser
• expansion device
• evaporator
• refrigerant
• thermostat

The thermostat controls operation so the compressor does not need to run continuously.

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Freezers

Freezers operate on the same basic principle as refrigerators.

However, they maintain:

lower temperatures

This usually requires the refrigeration system to operate across a larger temperature difference.

That can reduce COP and increase energy consumption.

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Air Conditioners

Air conditioners remove thermal energy and often moisture from indoor air.

The indoor evaporator:

absorbs thermal energy

The outdoor condenser:

releases thermal energy

The compressor requires electrical work.

Thus:

air conditioner = refrigeration system for a building

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Dehumidification

When warm humid air passes over a cold evaporator coil, water vapor can:

condense

The phase change:

gas → liquid

removes water from the air.

Therefore air conditioners often:

• cool the air
• reduce humidity

This can improve comfort even without an enormous temperature decrease.

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Domestic Heat Pumps

Heat pumps can provide:

• space heating
• space cooling
• water heating

Common types include:

• air-source heat pumps
• ground-source heat pumps
• water-source heat pumps

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Air-Source Heat Pumps

An air-source heat pump exchanges thermal energy with outdoor air.

During heating:

outdoor air → refrigerant → indoor air

During cooling:

indoor air → refrigerant → outdoor air

These systems are widely used because installation is relatively straightforward compared with ground-source systems.

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Ground-Source Heat Pumps

A ground-source heat pump exchanges thermal energy with the ground.

Underground temperatures often vary less throughout the year than outdoor air temperatures.

This can allow a smaller operating temperature difference and potentially improve performance.

However, installation can be more complex and expensive.

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Heat-Pump Water Heaters

A heat-pump water heater uses a refrigeration cycle to transfer thermal energy:

surrounding air → water

Rather than generating all the heating directly from electrical resistance, it moves existing thermal energy into the water.

This can reduce electrical energy consumption.

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Refrigerated Transport

Refrigeration systems are also used in:

• refrigerated trucks
• shipping containers
• food warehouses
• supermarket displays
• vaccine storage
• medical storage

All use the same fundamental idea:

use work to transfer thermal energy from cold to warm

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Refrigerator Door Open in a Room

Suppose you leave a refrigerator door open in a sealed room.

Will the refrigerator cool the room?

No.

The refrigerator removes Q_c from nearby air but releases:

Q_h = Q_c + W

back into the same room.

Since:

Q_h > Q_c

the net effect is:

room warms slightly

because the electrical work ultimately becomes thermal energy.

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Worked Example: Open Refrigerator

Suppose the refrigerator removes:

600 J

from nearby air.

The compressor uses:

200 J

of electrical work.

It releases:

Q_h = 600 + 200

Q_h = 800 J

to the room.

Net effect on room:

800 − 600

= 200 J added

Therefore the room becomes warmer overall.

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Why the Refrigerator Interior Gets Cold

The refrigerator does not "produce cold."

Instead, it:

removes thermal energy

from its interior.

The interior temperature falls because its internal energy decreases.

It is therefore more accurate to say:

the refrigerator transfers heat out

rather than:

the refrigerator creates cold

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Insulation

Refrigerators and buildings use insulation to reduce unwanted heat transfer.

For a refrigerator:

warm room → cold interior

is the natural direction of heat flow.

Insulation slows this transfer.

Better insulation means the compressor needs to operate less often.

Therefore:

better insulation → lower energy consumption

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Door Seals

Damaged refrigerator door seals allow warm room air to enter.

This increases:

• heat transfer into refrigerator
• compressor operating time
• electrical energy consumption

Maintaining good seals therefore improves practical efficiency.

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Refrigerator Placement

A refrigerator must release Q_h through its condenser.

If airflow around the condenser is poor, heat transfer becomes more difficult.

The compressor may then need to work harder.

Therefore adequate ventilation around refrigeration equipment can improve performance.

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Real Refrigerators and Irreversibility

Real refrigeration systems contain:

• compressor friction
• electrical resistance
• pressure losses
• heat transfer across finite temperature differences
• refrigerant flow resistance

These processes produce entropy.

Therefore:

COP_real < COP_ideal

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Refrigerator vs Heat Pump

The underlying physical device can be almost identical.

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Refrigerator vs Air Conditioner

A refrigerator and air conditioner are also fundamentally similar.

Refrigerator

Cools:

small insulated compartment

Air conditioner

Cools:

room or building

Both transfer thermal energy from a colder region to warmer surroundings using external work.

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Heat Pump vs Electric Heater

Resistance heater

Electrical energy is directly converted into thermal energy.

Approximately:

1 J electrical → 1 J heat

Heat pump

Electrical energy operates the compressor while additional energy is transferred from the environment.

For example:

1 J electrical + 3 J environment → 4 J indoor heating

Therefore the heat pump may have:

COP > 1

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COP Is Not Efficiency

This distinction is important.

Efficiency is generally:

useful energy output / energy input

and cannot exceed 1 for an ordinary energy-conversion device.

COP measures:

desired thermal-energy transfer / work input

Because much of the transferred energy comes from another reservoir rather than being created from the work input:

COP can exceed 1

There is no violation of energy conservation.

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Worked Example: Complete Refrigerator Problem

A refrigerator removes:

Q_c = 2400 J

from its interior and requires:

W = 800 J

Find Q_h

Q_h = Q_c + W

Q_h = 2400 + 800

Q_h = 3200 J

Find COP_R

COP_R = Q_c/W

COP_R = 2400/800

COP_R = 3.0

Therefore:

• energy removed = 2400 J
• work input = 800 J
• energy released = 3200 J
• refrigerator COP = 3.0

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Worked Example: Heat Pump

A heat pump has:

COP_HP = 4.5

and uses:

W = 2000 J

Find Q_h.

COP_HP = Q_h/W

Therefore:

Q_h = COP_HP × W

Q_h = 4.5 × 2000

Q_h = 9000 J

Now find Q_c:

Q_c = Q_h − W

Q_c = 9000 − 2000

Q_c = 7000 J

The heat pump transfers 7000 J from outside and uses 2000 J of work to deliver 9000 J indoors.

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Worked Example: Finding Work

A refrigerator removes:

3600 J

and has:

COP_R = 4.0

Use:

COP_R = Q_c/W

Rearrange:

W = Q_c/COP_R

W = 3600/4

W = 900 J

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Worked Example: Comparing Refrigerators

Refrigerator A:

Q_c = 1200 J

W = 400 J

COP_A = 1200/400

COP_A = 3

Refrigerator B:

Q_c = 1600 J

W = 400 J

COP_B = 1600/400

COP_B = 4

Therefore:

Refrigerator B has the higher COP

because it removes more thermal energy for the same work input.

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Common Mistakes

Mistake: Saying refrigerators create cold.
They remove thermal energy from a colder region.

Mistake: Saying heat naturally moves from cold to hot.
External work is required for sustained cold-to-hot transfer.

Mistake: Forgetting the compressor requires energy.
The compressor supplies the work needed to drive the cycle.

Mistake: Saying Q_h = Q_c.
Actually, Q_h = Q_c + W.

Mistake: Confusing refrigerator and heat-pump COP.
For a refrigerator, COP_R = Q_c/W. For a heat pump, COP_HP = Q_h/W.

Mistake: Saying COP > 1 violates energy conservation.
COP measures transferred energy relative to work input, not energy created.

Mistake: Treating COP as percentage efficiency.
A COP of 4 is normally written as 4, not 400% efficiency.

Mistake: Saying a refrigerator with its door open cools a closed room.
It adds net thermal energy to the room because of compressor work.

Mistake: Saying cold outdoor air contains no thermal energy.
Matter above absolute zero contains internal energy.

Mistake: Saying heat pumps and refrigerators use completely different principles.
They use essentially the same thermodynamic cycle but have different desired outputs.

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Key Terms

Refrigerator — device using work to remove thermal energy from a colder region and transfer it to warmer surroundings.

Heat pump — device using work to transfer thermal energy into a warmer region for heating.

Refrigerant — working fluid circulating through a refrigeration or heat-pump system.

Evaporator — component where refrigerant absorbs thermal energy and evaporates.

Compressor — component that performs work on refrigerant, increasing its pressure and temperature.

Condenser — component where refrigerant releases thermal energy and condenses.

Expansion valve — device that lowers refrigerant pressure before it enters the evaporator.

Cold reservoir — low-temperature region from which Q_c is removed.

Hot reservoir — high-temperature region receiving Q_h.

Coefficient of performance (COP) — ratio of desired thermal-energy transfer to required work input.

COP_R — refrigerator coefficient of performance, Q_c/W.

COP_HP — heat-pump coefficient of performance, Q_h/W.

Vapor-compression cycle — common refrigeration cycle using evaporation, compression, condensation and expansion.

Air-source heat pump — heat pump exchanging energy with outdoor air.

Ground-source heat pump — heat pump exchanging energy with the ground.

Latent heat — energy transferred during a phase change without a temperature change at constant pressure for a pure substance.

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Key Takeaways

• Refrigerators and heat pumps transfer thermal energy from colder regions to warmer regions.
• Cold-to-hot heat transfer requires external work.
• Most household systems obtain this work from electrical energy.
• A refrigerator removes Q_c from its cold interior.
• Work W is supplied to the compressor.
• The condenser releases Q_h to the warmer surroundings.
• Energy conservation gives Q_h = Q_c + W.
• The evaporator absorbs thermal energy.
• The compressor raises refrigerant pressure and temperature.
• The condenser releases thermal energy.
• The expansion valve lowers refrigerant pressure and temperature.
• Refrigerators use phase changes of a refrigerant to transport thermal energy effectively.
• A refrigerator's useful effect is removing Q_c.
• A heat pump's useful effect is delivering Q_h.
• Refrigerators and heat pumps therefore use essentially the same physics.
• Air conditioners operate as refrigeration systems for buildings.
• Reversible heat pumps can provide both heating and cooling.
• Refrigerator performance is measured using COP_R = Q_c/W.
• Heat-pump performance is measured using COP_HP = Q_h/W.
• For the same device, COP_HP = COP_R + 1.
• COP can be greater than 1 without violating energy conservation.
• COP is not the same as efficiency.
• Smaller temperature differences generally allow higher COP values.
• Ideal reversible COP values depend on the hot and cold reservoir temperatures.
• Real devices have lower COP because of irreversibility.
• Heat pumps can deliver several joules of thermal energy for each joule of electrical work consumed.
• Household applications include refrigerators, freezers, air conditioners, space-heating heat pumps and heat-pump water heaters.
• Leaving a refrigerator door open cannot cool a closed room; it causes net heating.
• Insulation reduces unwanted thermal-energy transfer and lowers refrigeration energy use.
• Refrigerators and heat pumps obey both the First and Second Laws of Thermodynamics.

Check Your Understanding

1. What is the main purpose of a refrigerator?
2. Why is work required to operate a refrigerator?
3. What do Q_c, Q_h and W represent?
4. Write the energy equation for a refrigerator.
5. Name the four main components of a vapor-compression refrigeration system.
6. What happens to the refrigerant in the evaporator?
7. What does the compressor do to the refrigerant?
8. Why can the refrigerant release energy to the room after compression?
9. What happens in the condenser?
10. What is the purpose of the expansion valve?
11. Explain the main difference between a refrigerator and a heat pump.
12. Write the equation for refrigerator COP.
13. Write the equation for heat-pump COP.
14. A refrigerator removes 1500 J while requiring 500 J of work. Calculate its COP.
15. For Question 14, calculate the energy released to the room.
16. A heat pump has COP_HP = 4 and uses 1500 J of electrical work. Calculate the energy delivered to the building.
17. Explain why a COP greater than 1 does not violate energy conservation.
18. Explain why a heat pump generally has a lower COP when the outdoor temperature becomes much colder.
19. Predict what happens to the temperature of a closed room if a refrigerator is operated with its door open. Explain your answer.
20. Compare a refrigerator, heat pump and heat engine in terms of the directions of heat transfer and work.