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Refrigerators and heat pumps

Work moving heat uphill: the coefficients of performance of refrigerators and heat pumps, their entropy limits, and the EER, SEER and ton ratings of American appliances.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to budget the energy of a refrigerator or heat pump, find its coefficient of performance and its Carnot limit, relate the two coefficients, and read the EER and ton ratings on an air conditioner.

2. What you already have

You know the energy and entropy budget of a heat engine and the Carnot limit $1 - T_c/T_h$. You know that heat flows spontaneously only from hot to cold. This lesson runs the engine the other way, using work to push heat uphill, and finds the limit on how cheaply that can be done.

3. Words for this lesson

4. Work can move heat uphill, and entropy prices it

A refrigerator takes heat $Q_c$ from a cold space at $T_c$, uses work $W$, and rejects heat $Q_h$ to a warmer space at $T_h$. The first law for a cycle gives

$$Q_h = Q_c + W.$$

What you want is $Q_c$ and what you pay for is $W$, so a refrigerator is rated by its coefficient of performance, $\text{COP}_R = Q_c/W$. Unlike an efficiency it can exceed one: moving heat can cost much less work than the heat moved.

The second law sets the limit. The cold space loses entropy $Q_c/T_c$; the device returns to its starting state each cycle; so the hot space must gain at least that much: $Q_h/T_h \geq Q_c/T_c$. Combining with the first law,

$$\text{COP}_R = \frac{Q_c}{Q_h - Q_c} \leq \frac{T_c}{T_h - T_c}.$$

A heat pump is the same machine valued for the heat it delivers indoors, with $\text{COP}_{\text{HP}} = Q_h/W \leq T_h/(T_h - T_c)$. Because $Q_h = Q_c + W$, the two coefficients differ by exactly one.

The limits depend on the difference of the temperatures in the denominator: pumping heat across a small difference is cheap, across a large one expensive. A kitchen refrigerator holding $4$ °C in a $22$ °C room could in principle move fifteen joules of heat for every joule of work. With no work at all, the COP would be infinite and heat would flow from cold to hot on its own — which is the Clausius statement of the second law saying it cannot.

5. The method, step by step, and how to check it

1. Decide what is useful. For a refrigerator or air conditioner, the heat removed from the cold space; for a heat pump, the heat delivered to the warm space. The work is always what is paid for.
2. Budget the energy with $Q_h = Q_c + W$, per cycle or per second.
3. Compute the coefficient from its definition.
4. Compute the limit with absolute temperatures, and remember the difference in the denominator is the same in kelvin and Celsius: for $-18$ °C inside and $25$ °C outside, $T_c/(T_h - T_c) = 255/43 = 5.9$.
5. Convert ratings when they come in American units: an EER in BTU/h per watt divided by $3.412$ is a COP, and a ton of cooling is $12\,000$ BTU/h, about $3.52$ kW.

Checks. A real device's coefficient is below its limit, typically a third to a half of it. A heat pump's coefficient is always at least one, since it delivers at least its own work as heat. The coefficient falls as the temperature difference grows, which is why heat pumps struggle in the coldest weather and air conditioners work hardest on the hottest days. And the heat rejected is always larger than the heat removed.

6. Why a heat pump beats a furnace

An electric resistance heater turns each joule of electricity into one joule of heat, a heat pump's coefficient of performance of exactly one. A heat pump moving heat from outdoors at $0$ °C into a house at $20$ °C has a Carnot limit of $293/20 = 14.7$; real air-source units manage about $3$ at that temperature. Each joule of electricity then brings two more joules in from the cold outdoors, for three joules of heat delivered.

Even compared with burning gas directly, the heat pump can win. A power plant turning gas into electricity at $50$ percent efficiency, feeding a heat pump with a COP of $3$, delivers $1.5$ joules of heat for each joule of gas burned, against at most $0.95$ for a high-efficiency gas furnace. The second law does not forbid getting more heat than the energy you paid for, because most of that heat was already there outdoors; you paid only to lift it.

The trade-off appears in the cold. At $-15$ °C outdoors the limit falls to $293/35 = 8.4$ and real units drop toward $2$, so heat pumps sold for the northern United States are rated for the cold climate, and many keep a backup heater for the coldest nights.

7. Rating systems on American appliances

The Department of Energy's appliance labels express coefficients of performance in mixed units that are easy to misread. The EER of an air conditioner is its cooling in BTU per hour divided by its power in watts, measured at a single hot condition, $95$ °F outdoors; the SEER is the same ratio averaged over a cooling season. Since one watt is $3.412$ BTU per hour, an EER of $12$ is a COP of $12/3.412 = 3.5$. Federal minimums since 2023 require a SEER2 of about $13.4$ to $14.3$ for new central air conditioners, depending on the region.

Heat pumps also carry an HSPF, the heating season's BTU delivered per watt-hour, so an HSPF of $8.8$ is a seasonal COP of about $2.6$. Cooling capacity is sold in tons, a unit from the days of ice houses: one ton is the cooling of melting a ton of ice in a day, $12\,000$ BTU per hour. A typical house needs two to five tons.

8. The refrigeration cycle in a real appliance

Inside a household refrigerator a refrigerant, today usually R-600a (isobutane) or R-134a, circulates around a loop. The compressor squeezes the vapor, heating it above room temperature; in the coils on the back it gives up heat to the kitchen and condenses to a liquid. The liquid passes through a narrow expansion valve into the low-pressure coils inside the cabinet, where it partly evaporates and cools far below the food's temperature. There it absorbs heat as it finishes boiling, and the vapor returns to the compressor.

The cycle runs heat uphill by exploiting the latent heat of the refrigerant and the fact that a liquid's boiling point depends on pressure: low pressure lets it boil cold, inside; high pressure makes it condense hot, outside. Each temperature difference across a coil creates entropy, which is why real coefficients of performance sit well below the Carnot limit.

9. In the world: heat pumps in a cold climate

Heat pumps have become the fastest-growing way to heat American homes, and manufacturers now sell cold-climate models for states like Maine and Minnesota. The entropy limit shows both why they work and where they struggle.

A house kept at $20$ °C ($293$ K) on a mild $5$ °C ($278$ K) day has a heat-pump limit of $293/15 = 19.5$, and a good unit reaches a COP near $4$. On a $-20$ °C ($253$ K) day the limit falls to $293/40 = 7.3$, and the real COP to about $1.8$. A house losing $15$ kW of heat on that cold day then needs $15/1.8 = 8.3$ kW of electricity, against $15$ kW for resistance heating.

Maine's heat pump program, which passed its goal of $100\,000$ installations in 2023, relies on exactly this arithmetic: even at a COP of $2$, a heat pump halves the electricity needed for resistance heat, and over a whole winter, when most days are milder than the coldest, the seasonal average is closer to $3$. The limit also explains why radiators sized for a low water temperature help: delivering heat at $35$ °C instead of $60$ °C shrinks $T_h - T_c$ and raises the ceiling on performance.

10. In the world: the cost of keeping a warehouse frozen

Cold storage warehouses hold frozen food at about $-25$ °C, and their refrigeration plants are among the largest electricity users in the food industry. The Carnot limit sets the floor on that cost.

With the warehouse at $248$ K and heat rejected outdoors at $308$ K on a summer day, the limit is $248/60 = 4.1$. A warehouse that leaks $500$ kW of heat through its walls and doors, and from the goods and forklifts inside, needs at least $500/4.1 = 120$ kW of compressor power; real ammonia systems with a COP near $2$ need about $250$ kW, around $6$ GWh a year.

Every design choice follows from the formula. Thicker insulation reduces $Q_c$. Rejecting heat through evaporative condensers, which run cooler than dry coils, lowers $T_h$ and raises the limit. Holding the storage only as cold as the food needs raises $T_c$: at $-18$ °C instead of $-25$ °C the limit rises from $4.1$ to $4.8$, cutting the minimum power by about fifteen percent.

11. A refrigerator cannot cool a room

Leaving the refrigerator door open on a hot day feels as if it should help, since cold air spills out. But a refrigerator does not destroy heat; it moves it from inside the cabinet to the coils on the back, and adds the work its compressor consumes. With the door open, both the cabinet and the coils are in the kitchen. The heat taken from the open cabinet comes straight back out of the coils, along with every watt of electricity the compressor draws. The kitchen gets warmer, not cooler. An air conditioner works only because its hot side is outdoors.

A second confusion is to treat a coefficient of performance above one as a violation of energy conservation, as if a heat pump made energy. It delivers more heat than the electricity it uses because most of that heat was already outdoors; the electricity pays only for lifting it to a higher temperature. The first law holds exactly: heat delivered equals heat taken in plus work.

12. The limit for a kitchen refrigerator

1. A refrigerator holds $4$ °C in a kitchen at $22$ °C. Convert the inside temperature to kelvin.

   $T_c = 4 + 273 = 277\ \text{K}$

   The limit needs an absolute temperature on top.

2. Convert the kitchen temperature.

   $T_h = 22 + 273 = 295\ \text{K}$

   The coils reject heat to the kitchen.

3. Find the temperature difference.

   $T_h - T_c = 295 - 277 = 18\ \text{K}$

   The height the heat must be lifted.

4. Write the limit and substitute.

   $\text{COP}_{\max} = \dfrac{T_c}{T_h - T_c} = \dfrac{277}{18}$

   From the entropy balance.

5. Evaluate and interpret.

   $\text{COP}_{\max} \approx 15.4$

   A perfect refrigerator would move fifteen joules out for each joule of work; real ones manage two to four, because their coils run much colder and hotter than the air they touch.

13. A heat pump warming a house

1. A heat pump delivers heat to radiators at $35$ °C while drawing it from outdoor air at $-5$ °C. Convert both temperatures.

   $T_h = 308\ \text{K}, \qquad T_c = 268\ \text{K}$

   The heat pump works between the outdoor air and the water it heats.

2. Find the Carnot limit for the heat pump.

   $\text{COP}_{\max} = \dfrac{T_h}{T_h - T_c} = \dfrac{308}{40} = 7.7$

   A heat pump is valued for the heat it delivers.

3. Take a real coefficient of $3.0$ and find the work for $10$ kW of heat.

   $W = \dfrac{Q_h}{\text{COP}} = \dfrac{10}{3.0} = 3.33\ \text{kW}$

   The electrical power the compressor draws.

4. Find the heat taken from the outdoor air.

   $Q_c = Q_h - W = 10 - 3.33 = 6.67\ \text{kW}$

   Two thirds of the heat delivered comes from the cold outdoors.

5. Compare with an electric resistance heater.

   $W_{\text{resistance}} = 10\ \text{kW}$

   The heat pump uses a third of the electricity for the same heat.

6. Check the entropy balance.

   $\dfrac{Q_h}{T_h} - \dfrac{Q_c}{T_c} = \dfrac{10\,000}{308} - \dfrac{6670}{268} = 32.5 - 24.9 = 7.6\ \text{W/K} > 0$

   Positive, as a real device must be; a perfect one would create no entropy.

14. Reading an air conditioner's rating

1. A central air conditioner provides $3$ tons of cooling. Convert to BTU per hour.

   $3 \times 12\,000 = 36\,000\ \text{BTU/h}$

   One ton of cooling is $12\,000$ BTU per hour.

2. Convert to watts.

   $\dfrac{36\,000}{3.412} \approx 10\,550\ \text{W}$

   One watt is $3.412$ BTU per hour.

3. Find the electrical power at an EER of $12$.

   $P = \dfrac{36\,000}{12} = 3000\ \text{W}$

   EER is BTU per hour per watt.

4. Find its coefficient of performance.

   $\text{COP} = \dfrac{10\,550}{3000} \approx 3.5 = \dfrac{12}{3.412}$

   Both routes agree.

5. Find the heat rejected outdoors.

   $Q_h = 10\,550 + 3000 = 13\,550\ \text{W}$

   The outdoor unit rejects the house's heat plus the compressor's work.

6. Find the Carnot limit for $24$ °C inside and $35$ °C outside.

   $\dfrac{T_c}{T_h - T_c} = \dfrac{297}{11} = 27$

   The temperature difference is small, so the limit is high.

7. Compare the real coefficient with the limit.

   $\dfrac{3.5}{27} \approx 0.13$

   Real units fall far short because their refrigerant must be much colder than the indoor air and much hotter than the outdoor air to move heat quickly.

8. Find the running cost for eight hours at $15$ cents per kilowatt-hour.

   $3.0\ \text{kW} \times 8\ \text{h} \times \$0.15 = \$3.60$

   A realistic daily cost for a house in summer.

15. Your turn: a freezer holds $-18$ °C in a kitchen at $25$ °C. What is its greatest possible coefficient of performance?

1. Convert both temperatures to kelvin.

   $T_c = 255\ \text{K}, \qquad T_h = 298\ \text{K}$

   Add $273$ to each.

2. Find the difference.

   $T_h - T_c = 298 - 255 = 43\ \text{K}$

   The same in kelvin and Celsius.

3. Your turn: work this step out. Its working is at the end of the packet.

   Divide the two numbers.

16. Guided practice

On a hot day someone leaves a refrigerator door open to cool a sealed, insulated kitchen. The refrigerator draws $100$ W and removes heat from its interior at $200$ W. What happens to the kitchen?

• It warms, gaining $100$ W net
• It cools, losing $200$ W
• Its temperature does not change
• It cools until the refrigerator overheats

17. Guided practice

Complete the worked solution: what is the greatest coefficient of performance of a refrigerator holding $-3$ °C in a room at $27$ °C?

1. Add $273$ to the inside temperature.

   $T_c = -3 + 273 =$ _c K

   The limit uses absolute temperatures.

2. Subtract the inside temperature from the room temperature.

   $T_h - T_c = 27 - (-3) =$ _d K

   A temperature difference is the same in either scale.

3. Divide the cold temperature by the difference.

   $\text{COP}_{\max} = \dfrac{T_c}{T_h - T_c} =$ _p

   The entropy removed from inside must at least be delivered to the room.

18. Guided practice

A device moves heat from a cold side at $265$ K to a hot side at $T_h$. Match each quantity to its formula.

19. Practice

Three refrigerators each run for a minute. Fill in the heat each rejects and its coefficient of performance.

20. Practice

A cold store is kept at $-73$ °C in surroundings at $27$ °C. What is the greatest coefficient of performance its refrigeration unit could have?

Answer:

21. Practice

A heat pump delivers $9$ kW of heat to a house with a coefficient of performance of $2.5$. How much electrical power does it draw?

Answer: unit: W / MW / kW

22. Somewhere new

A window air conditioner is rated at $18000$ BTU per hour of cooling with an energy efficiency ratio (EER) of $12$. What electrical power does it draw when running?

Answer: unit: W / MW / kW

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Three refrigerators each run for a minute. Fill in the heat each rejects and its coefficient of performance.

25. What you can do now

You can find how much work it takes to move heat from cold to hot. Explain to someone why leaving the refrigerator door open warms the kitchen.

Working for the steps left to you

15. Your turn: a freezer holds $-18$ °C in a kitchen at $25$ °C. What is its greatest possible coefficient of performance?, step 3