Refrigerators and heat pumps — COP: Exercises — Physics deep dive

Here $Q_{C}$ is the heat pulled from the cold reservoir (magnitude, in joules), $Q_{H}$ the heat dumped to the hot reservoir, $W_{in}$ the work paid in (electricity to the compressor). All three are positive numbers; directions are already baked into the formulas.

Sign-convention map (read this before the problems)

Every problem below hinges on knowing which direction $Q$ and $W$ point. The figure below fixes the rule once, as a picture: arrows into the working substance are positive inputs, arrows out are outputs, and the three always balance to zero over a cycle.

Figure — Refrigerators and heat pumps — COP

Quantity	Direction (for a fridge / heat pump)	Sign as used here
$Q_{C}$	heat in, pulled from the cold reservoir	$+$
$W_{in}$	work in, from the mains/compressor	$+$
$Q_{H}$	heat out, dumped to the hot reservoir	$+$ (leaves the substance)
balance	$Q_{C} + W_{in} = Q_{H}$ over one full cycle	$Δ U = 0$

The picture and the table say the same thing: two arrows in ( $Q_{C}$ , $W_{in}$ ), one arrow out ( $Q_{H}$ ), and they sum to the same total.

Level 1 — Recognition

L1.1

A machine takes in $Q_{C} = 300 J$ from a cold body and requires $W_{in} = 100 J$ of work per cycle. Which quantity is the "benefit" if the device is used as a fridge, and what is $COP_{r e f}$ ?

Recall Solution

What / why. A fridge is judged by the cooling it does, so the benefit is $Q_{C} = 300 J$ ; the cost is the work $W_{in} = 100 J$ . $COP_{r e f} = \frac{Q _{C}}{W _{in}} = \frac{300}{100} = 3.$ Meaning: every 1 J of electricity removes 3 J of heat from inside.

L1.2

Same machine as L1.1. Without recomputing from scratch, state its heat-pump COP.

Recall Solution

Why the shortcut works. The two COPs always differ by exactly 1 (same denominator $W_{in}$ , numerators differ by $Q_{H} - Q_{C} = W_{in}$ ). $COP_{h p} = COP_{r e f} + 1 = 3 + 1 = 4.$ Check via definition: $Q_{H} = Q_{C} + W_{in} = 300 + 100 = 400$ , so $COP_{h p} = Q_{H} / W_{in} = 400/100 = 4.$ ✓

L1.3

A fridge runs between inside $T_{C} = 250 K$ and room $T_{H} = 300 K$ . Write (do not evaluate yet) the formula for its maximum possible COP as a refrigerator, and say which temperature belongs on top.

Recall Solution

For a refrigerator the cold temperature sits on top (cold is the benefit): $COP_{r e f}^{C a r n o t} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{300 - 250} .$ The denominator is the climb $T_{H} - T_{C}$ .

Level 2 — Application

L2.1

A refrigerator with $COP_{r e f} = 5$ consumes $W_{in} = 120 J$ of work per cycle. Find $Q_{C}$ and $Q_{H}$ .

Recall Solution

Step 1 (cooling). $Q_{C} = COP_{r e f} \times W_{in} = 5 \times 120 = 600 J$ . Step 2 (energy balance, first law). $Q_{H} = Q_{C} + W_{in} = 600 + 120 = 720 J$ . The room receives 720 J while the inside loses only 600 J — the extra 120 J is the work, now warming the room.

L2.2

A Carnot heat pump keeps a house at $T_{H} = 293 K$ while the outside is $T_{C} = 273 K$ . Find $COP_{h p}$ , and how much heat it delivers per 1 J of electricity.

Recall Solution

Why heat-on-top. A heat pump's benefit is the heating $Q_{H}$ , so $T_{H}$ goes on top. $COP_{h p}^{C a r n o t} = \frac{T _{H}}{T _{H} - T _{C}} = \frac{293}{293 - 273} = \frac{293}{20} = 14.65.$ So each 1 J of electricity ideally delivers 14.65 J of heat into the house — vastly better than a 1:1 electric heater.

L2.3

An inventor claims a fridge running between $T_{C} = 280 K$ and $T_{H} = 310 K$ achieves $COP_{r e f} = 12$ . Is this allowed by the second law?

Recall Solution

Step 1 — the ceiling. $COP_{r e f}^{C a r n o t} = \frac{280}{310 - 280} = \frac{280}{30} = 9.33.$ Step 2 — compare. The claimed $12 > 9.33 = COP_{r e f}^{C a r n o t}$ . No real or reversible device can beat Carnot, so the claim is impossible — it would generate negative entropy in the universe. Reject.

Level 3 — Analysis

L3.1

A Carnot fridge holds $T_{C} = 250 K$ while the room warms from $T_{H} = 300 K$ to $T_{H} = 310 K$ . Compute $COP_{r e f}^{C a r n o t}$ before and after. Does raising the room temperature help or hurt, and why?

Recall Solution

Before: $\frac{250}{300 - 250} = \frac{250}{50} = 5.0$ . After: $\frac{250}{310 - 250} = \frac{250}{60} \approx 4.17$ . Why it drops. The denominator is the "hill height" $T_{H} - T_{C}$ . A hotter room makes the hill steeper, so pumping the same heat costs more work — COP falls. The plot below makes this visible: both COP curves slide down as the gap on the horizontal axis grows, and the two plum dots mark exactly this before/after pair.

The figure below traces how COP falls as the temperature gap widens; the two plum dots are this problem's before-and-after points (gaps of 50 K and 60 K). Read it as: push the reservoirs further apart and the leverage collapses.

L3.2

A fridge with its door left open sits in a sealed, insulated room. Over time, does the room get colder, warmer, or stay the same? Justify with $Q_{H}, Q_{C}, W_{in}$ .

Recall Solution

Reasoning. Each cycle the machine pulls $Q_{C}$ from the room-air (through the open door) and dumps $Q_{H} = Q_{C} + W_{in}$ back into the same room (through the coils). Net heat added to the room per cycle: $Q_{H} - Q_{C} = W_{in} > 0.$ The $Q_{C}$ cancels — it left the room and came right back. What remains is the compressor work $W_{in}$ , all converted to heat. The room gets warmer, at exactly the rate of electrical power drawn.

L3.3

Two Carnot refrigerators share the same room $T_{H} = 300 K$ . Fridge A cools to $T_{C} = 290 K$ ; Fridge B (a deep freezer) cools to $T_{C} = 250 K$ . Which has the higher COP, and by roughly what factor?

Recall Solution

$A: \frac{290}{300 - 290} = \frac{290}{10} = 29, B: \frac{250}{300 - 250} = \frac{250}{50} = 5.$ Fridge A's COP is $29/5 = 5.8 \times$ that of B. Why. Colder targets mean a bigger gap $T_{H} - T_{C}$ and a smaller numerator — both push COP down. Deep freezing is expensive.

Level 4 — Synthesis

L4.1

A real fridge achieves 40% of its Carnot COP. It runs between $T_{C} = 275 K$ and $T_{H} = 305 K$ and must remove $Q_{C} = 900 J$ per cycle. Find the actual COP, then the work $W_{in}$ it must draw, then $Q_{H}$ .

Recall Solution

Step 1 — Carnot ceiling. $COP_{r e f}^{C a r n o t} = \frac{275}{305 - 275} = \frac{275}{30} \approx 9.1 \overline{6}$ (the bar means $9.1666 \dots$ , exactly $\frac{55}{6}$ ). Step 2 — real COP. $COP_{r e a l} = 0.40 \times 9.1 \overline{6} = 3.6 \overline{6}$ (i.e. $3.6666 \dots = \frac{11}{3}$ ). Step 3 — work. $W_{in} = \frac{Q _{C}}{COP _{r e a l}} = \frac{900}{3.6 6} \approx 245.45 J$ . Step 4 — heat dumped. $Q_{H} = Q_{C} + W_{in} = 900 + 245.45 \approx 1145.45 J$ . The ideal machine would need only $900/9.1 \overline{6} \approx 98.2 J$ ; the real one needs $\approx 245$ J because of irreversibility.

L4.2

A device is used as a heat pump in winter and, in summer, the same unit runs in reverse as a fridge (air conditioner). In winter $T_{H} = 293 K$ , $T_{C} = 273 K$ . Find its Carnot heat-pump COP; then confirm the matching Carnot refrigerator COP two independent ways.

Recall Solution

Heat-pump COP: $\frac{293}{293 - 273} = \frac{293}{20} = 14.65$ . Way 1 — subtract 1: $COP_{r e f} = COP_{h p} - 1 = 14.65 - 1 = 13.65$ . Way 2 — direct formula: $\frac{T _{C}}{T _{H} - T _{C}} = \frac{273}{20} = 13.65$ . ✓ Both agree, confirming $COP_{h p} = COP_{r e f} + 1$ is just algebra on the same $W_{in}$ .

L4.3

A Carnot heat pump delivers $Q_{H} = 1200 J$ per cycle to a house at $T_{H} = 300 K$ , drawing heat from outside air at $T_{C} = 280 K$ . Find $W_{in}$ and the entropy change of the universe per cycle. Comment.

Recall Solution

Step 1 — COP. $COP_{h p} = \frac{300}{300 - 280} = \frac{300}{20} = 15$ . Step 2 — work. $W_{in} = \frac{Q _{H}}{COP _{h p}} = \frac{1200}{15} = 80 J$ . Step 3 — heats. $Q_{C} = Q_{H} - W_{in} = 1200 - 80 = 1120 J$ . Step 4 — entropy of universe. For a reversible (Carnot) cycle, entropy pulled from cold equals entropy dumped hot: $Δ S = \frac{Q _{H}}{T _{H}} - \frac{Q _{C}}{T _{C}} = \frac{1200}{300} - \frac{1120}{280} = 4.0 - 4.0 = 0.$ Comment. $Δ S_{u ni v er se} = 0$ is the signature of a reversible machine — it sits exactly on the second-law boundary. Any real unit would show $Δ S > 0$ and need more than 80 J.

Level 5 — Mastery

L5.1

A heat pump warms a house that leaks heat to the outdoors at a rate of $8000 J$ per second (8 kW). To hold steady temperature, the pump must deliver $Q_{H} = 8 kW$ . It runs at 50% of Carnot between $T_{C} = 268 K$ and $T_{H} = 298 K$ . Find the electrical power $W_{in}$ it draws. Compare to a 1:1 resistive heater.

Recall Solution

Step 1 — Carnot COP. $COP_{h p}^{C a r n o t} = \frac{298}{298 - 268} = \frac{298}{30} \approx 9.9 \overline{3}$ (bar: $9.9333 \dots = \frac{149}{15}$ ). Step 2 — real COP. $COP_{r e a l} = 0.50 \times 9.9 \overline{3} \approx 4.9 \overline{6}$ (i.e. $4.9666 \dots = \frac{149}{30}$ ). Step 3 — power drawn. Per second, $\dot{W}_{in} = \frac{Q _{H}}{COP _{r e a l}} = \frac{8000}{4.9 6} \approx 1610.7 W$ . Step 4 — compare. A resistive heater has COP $= 1$ , so it would draw the full $8000 W$ . The heat pump uses $\approx 1611 W$ — about 5× less electricity for the same warmth. The difference is the free heat $Q_{C}$ dragged in from the cold outdoors.

L5.2

Two engineers argue. Engineer X insists a fridge can be made with $COP_{r e f} = \infty$ "if we just build it perfectly." Engineer Y says impossible. Using the Carnot formula, state under what single physical condition $COP_{r e f} \to \infty$ , and why the second law forbids reaching it.

Recall Solution

$COP_{r e f}^{C a r n o t} = \frac{T _{C}}{T _{H} - T _{C}} \to \infty ⟺ T_{H} - T_{C} \to 0, i.e. T_{H} = T_{C} .$ Why the limit blows up. As the two reservoirs approach the same temperature, the "hill" vanishes — moving heat costs essentially no work. Why it's unreachable. If $T_{H} = T_{C}$ there is no temperature difference and no distinct "cold" and "hot" bodies to pump between; heat wouldn't need pumping at all. For any useful fridge you need $T_{H} > T_{C}$ , so the denominator is strictly positive and COP is finite. Engineer Y is right: $\infty$ is a limit, never an achievable value.

L5.3

A Carnot refrigerator ( $T_{C} = 260 K$ , $T_{H} = 300 K$ ) removes $Q_{C} = 780 J$ per cycle. Its exhaust heat $Q_{H}$ is fed as the input to a Carnot heat engine operating between the same $T_{H} = 300 K$ and a colder sink at $T_{L} = 200 K$ . Find the work the engine produces, and hence the net work cost of running the fridge after crediting the engine's output.

Recall Solution

Symbols first. The fridge consumes $W_{in}$ ; the engine produces $W_{o u t}$ . These are opposite in direction, so we keep them as separate symbols and only combine them at the very end.

Fridge side. $COP_{r e f} = \frac{260}{300 - 260} = \frac{260}{40} = 6.5$ . Work to run fridge: $W_{in} = \frac{Q _{C}}{COP _{r e f}} = \frac{780}{6.5} = 120 J$ . Exhaust: $Q_{H} = Q_{C} + W_{in} = 780 + 120 = 900 J$ .

Engine side. A Carnot engine taking $Q_{H} = 900 J$ at $300 K$ and rejecting to $200 K$ has efficiency $η = 1 - \frac{T _{L}}{T _{H}} = 1 - \frac{200}{300} = \frac{1}{3} .$ Work produced: $W_{o u t} = η Q_{H} = \frac{1}{3} \times 900 = 300 J$ .

Net cost. The engine hands back $W_{o u t} = 300 J$ of work while the fridge only needed $W_{in} = 120 J$ : $W_{n e t} = W_{in} - W_{o u t} = 120 - 300 = - 180 J .$ A negative cost means the combination produces surplus work — but note the engine dumps its own reject heat $Q_{H} - W_{o u t} = 900 - 300 = 600 J$ into the $200 K$ sink, which is a third reservoir. There's no free lunch: the surplus is paid for by heat flowing down to the very cold $200 K$ sink. (Uses Heat Engines and Efficiency and Carnot Cycle.)

Connections

Carnot Cycle — supplies every COP ceiling used above.
Second Law of Thermodynamics — why L2.3 and L5.2 fail; why COP is finite.
Entropy — the $Q / T$ bookkeeping behind L4.3.
Heat Engines and Efficiency — the reversed device in L5.3.
First Law of Thermodynamics — $Q_{H} = Q_{C} + W_{in}$ used throughout.