Refrigerators and heat pumps — COP: Worked examples

Before we start, the four tools we reuse (each already earned in the parent):

The scenario matrix

Every exam question on COP falls into exactly one of these cells (A–H). The map below groups them by which move solves them — the same two-move routing the mnemonic at the bottom uses. The worked examples hit all eight cells.

#	Cell (case class)	What's tricky	Example
A	Given $T_{H}, T_{C}$ → find Carnot COP	plain plug-in, kelvin	Ex 1
B	Celsius given → must convert	the °C→K trap	Ex 2
C	Non-ideal device: COP & $W$ given → find $Q_{C}, Q_{H}$	first law, not Carnot	Ex 3
D	Heat-pump viewpoint of the same machine	$+ 1$ relation, "beats a heater"	Ex 4
E	Limiting case: $T_{C} \to T_{H}$ (gap → 0)	COP → ∞, why	Ex 5
F	Degenerate case: $T_{C} \to 0$ K or huge gap	COP → 0, third-law hint	Ex 6
G	Door-open fridge (real-world twist)	net heating of a room	Ex 7
H	Real vs ideal ratio (exam twist)	fraction of Carnot achieved	Ex 8

Figure — Refrigerators and heat pumps — COP

The figure plots the Carnot fridge COP on the vertical axis (labelled $COP_{r e f}^{C}$ , on a log scale so the huge and tiny values both fit) against the horizontal axis — the temperature gap $T_{H} - T_{C}$ in kelvin, for a fixed hot side $T_{H} = 300 K$ . Interrogate it as you read:

Far left (blue curve rockets up): the gap is $\approx 1 K$ , so the denominator is tiny and COP explodes past 200 — this is exactly the green dot Ex 5 ( $T_{C} \to T_{H}$ , COP $\to \infty$ ).
Middle: the white Ex 8 dot sits at gap $30 K$ , COP $= 9$ ; the yellow Ex 1 dot at gap $50 K$ , COP $= 5$ . Notice the curve is already dropping steeply here.
The dashed white line at COP $= 1$ : below it, a fridge moves less heat than the work it pays. Only the far-right red Ex 6 dot (gap $297 K$ ) falls below this line, at COP $\approx 0.01$ .
Reading the slope: every doubling of the gap roughly halves the COP — the "hill" grows and the leverage collapses. That single trend explains cells A, E, F and H at once.

Cell A — plain Carnot plug-in

Worked example Ex 1 · Freezer against a warm room

A freezer keeps its inside at $T_{C} = 250 K$ while the kitchen is $T_{H} = 300 K$ . What is the maximum possible $COP_{r e f}$ ?

Forecast: the gap is $50 K$ — not tiny, not huge. Guess: COP a bit above 4? Write your guess down.

Identify the two temperatures: $T_{C} = 250$ , $T_{H} = 300$ (already kelvin). Why this step? Carnot COP is built from entropy $Q / T$ , which only makes sense on the absolute kelvin scale — check the units before touching the formula.
Compute the climb $T_{H} - T_{C} = 300 - 250 = 50 K$ . Why this step? The denominator is always the hill the heat must be pumped up; isolating it first prevents arithmetic slips.
Apply $COP_{r e f}^{C} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{50} = 5$ . Why this step? This is the ceiling — no real freezer with these temperatures can beat COP $= 5$ .

Verify: the fridge formula puts the cold temperature $T_{C}$ on top (benefit = cooling), giving $250/50 = 5$ ; the heat-pump twin instead puts the hot temperature $T_{H}$ on top, $COP_{h p} = \frac{T _{H}}{T _{H} - T _{C}} = \frac{300}{50} = 6 = 5 + 1$ ✓ (the $+ 1$ relation holds).

Cell B — the Celsius trap

Worked example Ex 2 · Air-conditioner quoted in °C

An AC cools a room to $T_{C} = 20 ° C$ while dumping heat outside at $T_{H} = 40 ° C$ . Find the maximum $COP_{r e f}$ . A student writes $\frac{20}{40 - 20} = 1.0$ . Is that right?

Forecast: the naive Celsius answer is $1.0$ . But a $20 K$ gap on a $\sim 300 K$ absolute scale is small, so the true COP should be much larger than 1. Predict: around 14–15?

Convert both to kelvin using the exact offset $273.15$ : $T_{C} = 20 + 273.15 = 293.15 K$ , $T_{H} = 40 + 273.15 = 313.15 K$ . Why this step? $Q \propto T$ came from entropy $Q / T$ ; ratios of Celsius numbers are meaningless because °C has an arbitrary zero. See Entropy. (Rounding to $+ 273$ is fine for rough work, but a strict thermodynamics context uses $273.15$ .)
Climb: $T_{H} - T_{C} = 313.15 - 293.15 = 20 K$ . (Note: the gap is the same in °C or K — but the numerator is not!) Why this step? Temperature differences are identical in both scales; temperature values are not — that's the whole trap.
$COP_{r e f}^{C} = \frac{293.15}{20} = 14.66$ . Why this step? This is the honest answer; the student's $1.0$ was off by a factor of ~15.

Verify: the correct ratio uses the gap $20$ with the absolute numerator $293.15$ , giving $14.66$ , over 14× the wrong Celsius answer. A tiny hill on a large absolute floor ⇒ big COP, consistent with Ex 1's trend.

Cell C — non-ideal device, find the heats

Worked example Ex 3 · A real fridge, COP and work given

A real refrigerator has $COP_{r e f} = 3.5$ and its compressor does $W = 120 J$ of work per cycle. Find $Q_{C}$ (heat pulled from inside) and $Q_{H}$ (heat dumped to the room).

Forecast: benefit ÷ cost $= 3.5$ , so $Q_{C}$ should be $3.5 \times$ the work $\approx 420 J$ , and $Q_{H}$ a bit more. Guess $Q_{H} \approx 540 J$ ?

From the definition $COP_{r e f} = \frac{Q _{C}}{W}$ , solve $Q_{C} = COP_{r e f} \times W = 3.5 \times 120 = 420 J$ . Why this step? We were given the actual COP, so we use the general definition (not Carnot) — this holds for any device, ideal or not.
Use the first law $Q_{H} = Q_{C} + W = 420 + 120 = 540 J$ . Why this step? Over one full cycle the working substance returns to its starting state, so its internal energy is unchanged — "heat in equals heat out" — which is precisely why the balance $Q_{H} = Q_{C} + W$ holds no matter how good the fridge is. See First Law of Thermodynamics.

Verify: units all joules ✓. The room gets $540 J$ while the inside only loses $420 J$ — the missing $120 J$ is exactly the electricity you paid, now turned to heat. $Q_{H} > Q_{C}$ ✓ (surplus $= W$ ).

Cell D — same machine, heat-pump viewpoint

Worked example Ex 4 · Heat pump vs a plain electric heater

The exact machine from Ex 3 is now sold as a heat pump to warm a house. Each cycle it still does $W = 120 J$ and delivers $Q_{H} = 540 J$ . What is $COP_{h p}$ , and how does it compare to a $\$$1-per-joule electric heater (COP$ =1$)?

Forecast: heating COP should be fridge COP plus one, so $3.5 + 1 = 4.5$ . That means ~4.5× the heat of a plain heater for the same electricity.

Use $COP_{h p} = \frac{Q _{H}}{W} = \frac{540}{120} = 4.5$ . Why this step? For a heat pump the benefit is the heat delivered $Q_{H}$ , not the cooling — different numerator, same cost $W$ .
Cross-check with the relation $COP_{h p} = COP_{r e f} + 1 = 3.5 + 1 = 4.5$ ✓. Why this step? Confirms we didn't mix up numerators; both COPs share denominator $W$ and their numerators differ by $Q_{H} - Q_{C} = W$ .

Verify: a plain electric heater turns $120 J$ of electricity into $120 J$ of heat (COP $= 1$ ). This pump delivers $540 J$ — 4.5× more — because it also drags free heat in from outside. That extra is why heat pumps save money. See Heat Engines and Efficiency.

Cell E — limiting case: the gap shrinks to zero

Worked example Ex 5 · What happens as

$T_{C} \to T_{H}$ ? A heat pump warms a cellar that is already almost as warm as outside: $T_{H} = 291 K$ , $T_{C} = 290 K$ . Then imagine $T_{C} \to 291$ . What does $COP_{h p}^{C}$ do?

Forecast: the hill is nearly flat, so pumping heat should be nearly free → COP enormous, heading to infinity. Guess: 291?

Climb $= T_{H} - T_{C} = 291 - 290 = 1 K$ . Why this step? A $1 K$ denominator is what makes the ratio explode; isolating it shows the mechanism.
$COP_{h p}^{C} = \frac{291}{1} = 291$ . Why this step? Almost no work needed — you're barely lifting heat at all.
Take the limit $T_{C} \to T_{H}$ : denominator $\to 0^{+}$ , numerator stays $T_{H}$ , so $COP_{h p}^{C} \to + \infty$ . Why this step? This is the left end of the curve in the figure — no gap means no hill means (ideally) infinite leverage. The Second Law of Thermodynamics still forbids a literal zero-work transfer, so infinity is only the mathematical ceiling.

Verify: on the figure, as the gap → 0 the blue curve shoots up ✓. Fridge twin: $COP_{r e f}^{C} = 290/1 = 290 = 291 - 1$ ✓.

Cell F — degenerate case: cold reservoir near absolute zero

Worked example Ex 6 · Trying to reach very low temperatures

A cryo-fridge dumps to a room at $T_{H} = 300 K$ and must cool a sample to $T_{C} = 3 K$ . Find $COP_{r e f}^{C}$ . Then ask: what happens as $T_{C} \to 0$ ?

Forecast: huge gap ( $297 K$ ) with a tiny cold temperature on top → COP should be tiny, well below 1. Guess ~0.01?

Climb $= 300 - 3 = 297 K$ . Why this step? A near-maximal hill lives at the far-right of the curve — where COP is smallest.
$COP_{r e f}^{C} = \frac{3}{297} = 0.01010$ . Why this step? You must pay ~99 J of work for every 1 J of heat removed — extremely expensive cooling.
Limit $T_{C} \to 0 K$ : numerator $\to 0$ , denominator $\to T_{H}$ , so $COP_{r e f}^{C} \to 0$ . Why this step? Removing the last scraps of heat near absolute zero costs unbounded work per joule — a fingerprint of the third law. This is the right end of the figure's curve.

Verify: $3/297 \approx 0.0101$ , indeed $≪ 1$ — one of the rare cases where a refrigerator COP is below 1 ✓ (contrast the parent's "COP usually $> 1$ "). Heat-pump twin $= 0.0101 + 1 = 1.0101$ , still just above 1 ✓.

Cell G — the door-open fridge (real-world twist)

Worked example Ex 7 · Can a fridge cool your kitchen with its door open?

You leave a fridge running with its door open in a sealed kitchen. Per cycle it removes $Q_{C} = 300 J$ from the (now open) inside and its motor uses $W = 90 J$ . Does the kitchen get cooler, warmer, or stay the same?

Forecast: intuition says "cold air pours out → cooler." Physics says look at $Q_{H}$ dumped behind the fridge. Guess before reading step 2.

The heat dumped to the room is $Q_{H} = Q_{C} + W = 300 + 90 = 390 J$ per cycle. Why this step? First law again — every joule the fridge removes plus every joule of electricity ends up behind the coils as heat.
But with the door open, the $Q_{C} = 300 J$ it "removed" was taken from the same room it dumps into. Net heat added to the sealed kitchen per cycle $= Q_{H} - Q_{C} = W = 90 J$ . Why this step? The cooling and the dumping now happen in one connected space, so they cancel — except the paid work, which can't cancel.
Net effect: $+ 90 J$ of heat per cycle → the kitchen warms up. Why this step? An open-door fridge is a $90 J$ -per-cycle heater. Only a closed door (cold inside separated from the warm room) actually keeps food cold.

Verify: net heating $= W = 90 J > 0$ ✓. This matches the Second Law of Thermodynamics: you cannot cool a closed room by running a device inside it — the work you feed in has nowhere to go but heat.

Cell H — real fraction of Carnot (exam twist)

Worked example Ex 8 · How good is my real fridge vs the ideal?

A fridge operates between $T_{C} = 270 K$ and $T_{H} = 300 K$ . Its measured $COP_{r e f} = 6.0$ . What fraction of the Carnot maximum does it achieve? If it moves $Q_{C} = 600 J$ per cycle, how much work does it actually pay, and how much would an ideal device pay?

Forecast: Carnot COP with a $30 K$ gap is $270/30 = 9$ . So the real one at 6 achieves about two-thirds of ideal, and pays more work than ideal for the same cooling.

Carnot ceiling: $COP_{r e f}^{C} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{270}{300 - 270} = \frac{270}{30} = 9.0$ . Why this step? This sets the best-possible bar for these two temperatures; any real device between the same reservoirs must have a COP below this ceiling — so the measured $6.0$ had better be less than $9.0$ (it is).
Fraction of ideal $= \frac{COP _{r e f}}{COP _{r e f}^{C}} = \frac{6.0}{9.0} = 0.6667$ , i.e. $66.7%$ of Carnot. Why this step? This "second-law efficiency" is the one-number summary of how close the real machine gets to perfection — the exam's favourite quantity.
Real work for $Q_{C} = 600 J$ : rearrange the definition $COP_{r e f} = \frac{Q _{C}}{W}$ to get $W_{r e a l} = \frac{Q _{C}}{COP _{r e f}} = \frac{600}{6.0} = 100 J$ . Why this step? We are told the actual COP and the actual cooling, so the definition (not Carnot) gives the true electricity paid.
Ideal work for the same $600 J$ : $W_{i d e a l} = \frac{Q _{C}}{COP _{r e f}^{C}} = \frac{600}{9.0} = 66.7 J$ . Why this step? A perfect Carnot device moves the same heat for less work; the gap between $W_{r e a l}$ and $W_{i d e a l}$ is the energy wasted to irreversibility. See Carnot Cycle.

Verify: $6 < 9$ so real $<$ Carnot ✓ (never violated). Real work $100 J >$ ideal $66.7 J$ ✓ — you pay a $33.3 J$ penalty per cycle for imperfection, exactly the $\frac{1}{0.6667} - 1 = 50%$ extra work the second-law efficiency predicts.

Recall Quick self-test across the matrix

Which cell asks about a fridge with its door open? ::: Cell G — net room heating equals the work $W$ . As the temperature gap shrinks to zero, Carnot COP does what? ::: Cell E — it diverges to $+ \infty$ . When can a refrigerator's COP legitimately drop below 1? ::: Cell F — a huge gap with a very cold $T_{C}$ (e.g. $T_{C} = 3 K$ , $T_{H} = 300 K$ gives $\approx 0.0101$ ). First step whenever temperatures are given in °C? ::: Convert to kelvin (Cell B): $T (K) = T (° C) + 273.15$ .

Connections

Parent: Refrigerators & Heat Pumps — COP — this page is its worked-example drill.
Carnot Cycle — supplies the ceiling used in Cells A, B, E, F, H.
First Law of Thermodynamics — the $Q_{H} = Q_{C} + W$ used in Cells C, D, G.
Second Law of Thermodynamics — why COP $\to \infty$ is only a limit and why an open-door fridge can't cool a room.
Entropy — the $Q / T$ bookkeeping behind the Celsius trap in Cell B.
Heat Engines and Efficiency — contrasts leverage-ratio COP with fractional efficiency.