Refrigerators and heat pumps — COP: Question bank

Reminder of the symbols used everywhere below (all positive magnitudes per cycle):

$Q_{C}$ = heat pulled out of the cold reservoir (temperature $T_{C}$ ),
$Q_{H}$ = heat dumped into the hot reservoir (temperature $T_{H}$ ),
$W$ = work you pay in, with $Q_{H} = Q_{C} + W$ ,
COP = "benefit ÷ cost" — cooling $Q_{C} / W$ for a fridge, heating $Q_{H} / W$ for a heat pump.
Temperatures in these formulas are kelvin: $T (K) = T (° C) + 273$ .

True or false — justify

A fridge with its door open cools the room down over time.

False. It dumps

Q_{H} = Q_{C} + W

into the room while only removing

Q_{C}

from the air in front; the net effect is heating the room by exactly the work

W

the motor spends.

COP

can be less than 1.

True for a fridge — if

Q_{C} < W

(a big temperature hill),

COP_{r e f} = Q_{C} / W < 1

. But a heat pump can never be below 1 since

COP_{h p} = COP_{r e f} + 1 \geq 1

For fixed

T_{H}

and

T_{C}

, a real refrigerator can have a higher COP than a Carnot one.

False. The Carnot value

T_{C} / (T_{H} - T_{C})

is the ceiling set by the second law; any real device generates entropy, so its COP is strictly lower.

A heat pump and a refrigerator built from the same hardware always have COPs differing by exactly 1.

True. They share the same

W

; their numerators differ by

Q_{H} - Q_{C} = W

, so

COP_{h p} - COP_{r e f} = W / W = 1

exactly — independent of how ideal the machine is.

T_{C} \to T_{H}

(tiny gap), the maximum COP goes to infinity.

True. The denominator

T_{H} - T_{C} \to 0

, so

COP^{C a r n o t} \to \infty

— pumping heat across almost no hill costs almost no work. This never quite reaches infinity because a truly zero gap moves no heat usefully.

Increasing the work input

W

always increases

Q_{C}

False as a general law — it depends on the device. More

W

can move more heat, but for a fixed real machine the COP

= Q_{C} / W

is roughly a property of the temperatures, so

Q_{C}

scales with

W

only while COP stays fixed; you cannot exceed the Carnot ceiling by spending more.

A heat pump beats a 1:1 electric resistance heater.

True. The heater turns

W

into exactly

W

of heat (COP = 1). The heat pump delivers

Q_{H} = Q_{C} + W > W

, because it also drags free heat

Q_{C}

in from outside.

Q_{C} = 0

(nothing extracted from cold), the machine is a perfect heater.

Partly — with

Q_{C} = 0

you get

Q_{H} = W

, so all work becomes heat and

COP_{h p} = 1

. It is just a resistance heater; the "heat pump advantage" has vanished because nothing is being pumped uphill.

Spot the error

"COP must be under 1 because efficiency of an engine is under 1."

The error is treating COP like efficiency. Efficiency is useful work out ÷ heat in (a fraction of converted energy); COP is heat moved ÷ work paid — a leverage ratio that is usually greater than 1.

W = Q_{C} - Q_{H}

, so work equals the heat you took minus what you dumped."

Sign error. The device dumps more hot than it took cold, so

Q_{H} > Q_{C}

and the surplus is the paid work:

W = Q_{H} - Q_{C} > 0

. Writing

Q_{C} - Q_{H}

gives a negative "work", which is impossible for a driven pump.

"Carnot COP for a fridge at 4 °C in a 27 °C room is

4/ (27 - 4) \approx 0.17

Wrong scale. The

Q / T

entropy bookkeeping needs absolute temperature. Convert:

277/ (300 - 277) \approx 12

, not

0.17

. Celsius has an arbitrary zero, so ratios of °C are meaningless here.

"A reversible fridge has

Q_{C} / Q_{H} = T_{C} / T_{H}

, so it violates energy conservation because

Q_{C} < Q_{H}

No violation —

Q_{C} < Q_{H}

is required: the difference

Q_{H} - Q_{C} = W

is the work you paid in, exactly as the first law demands. Reversibility fixes the ratio; the first law supplies the missing

W

"Refrigerator and heat pump are the same machine, so they have the same COP."

They are the same hardware, but the benefit differs. The fridge cares about

Q_{C}

(cooling), the pump about

Q_{H}

(heating), giving different numerators — hence

COP_{h p} = COP_{r e f} + 1

, never equal.

"Since

Δ S_{u ni v er se} = 0

for a real fridge, we can use

Q_{C} / T_{C} = Q_{H} / T_{H}

directly."

Only a reversible fridge has

Δ S_{u ni v er se} = 0

. A real one generates entropy (

Δ S_{u ni v er se} > 0

), so

Q_{H} / Q_{C} > T_{H} / T_{C}

— more heat dumped hot per heat taken cold — meaning a real device wastes work and has a lower COP.

Why questions

Why does the second law forbid a fridge that needs zero work?

Zero-work heat transfer from cold to hot would move entropy

Q_{C} / T_{C}

off the cold body and dump only

Q_{C} / T_{H}

(smaller) on the hot, decreasing the universe's entropy — forbidden. You must pay

W

so the extra dumped heat carries the entropy balance.

Why is the denominator of both Carnot COPs the same

T_{H} - T_{C}

Both COPs equal (heat of interest)

/ W

, and for a reversible cycle

W = Q_{H} - Q_{C}

which the temperature relation turns into a factor of

(T_{H} - T_{C})

. The "climb"

T_{H} - T_{C}

is the common cost; only the top (which heat you value) changes.

Why does a bigger temperature gap make a fridge worse (lower COP)?

A bigger gap means a steeper hill to push heat up. Reversibility forces more dumped heat per unit taken, so more of your work is "spent on the climb" — the denominator

T_{H} - T_{C}

grows while the numerator shrinks, dropping the COP.

Why can we treat

Q_{H}

Q_{C}

W

as positive magnitudes without sign confusion?

Because we already fixed each arrow's direction physically (cold in, hot out, work in) in the energy diagram. Once directions are set by hand, the algebra

Q_{H} = Q_{C} + W

uses only sizes, avoiding a competing sign convention.

Why does

COP_{h p}

have a hard floor of 1 but

COP_{r e f}

does not?

COP_{h p} = Q_{H} / W = (Q_{C} + W) / W = COP_{r e f} + 1 \geq 1

; even if it pumps nothing (

Q_{C} = 0

), the work itself becomes delivered heat. A fridge only counts

Q_{C}

, which can shrink below

W

, so its COP can dip under 1.

Why is a heat pump's advantage largest in mild weather and smallest in extreme cold?

The COP is

T_{H} / (T_{H} - T_{C})

; a warmer outside (larger

T_{C}

, smaller gap) makes the denominator tiny and COP huge. In deep cold the gap widens, COP falls toward 1, and the pump barely beats a plain heater.

Edge cases

What happens to

COP_{r e f}

T_{C} \to 0

K (cold reservoir near absolute zero)?

COP_{r e f} = T_{C} / (T_{H} - T_{C}) \to 0

. Extracting the last bits of heat from something near absolute zero requires enormous work — an echo of the third law: you cannot reach 0 K with finite work.

W = 0

but

T_{C} < T_{H}

, what does the model predict?

With

W = 0

COP_{r e f} = Q_{C} /0

is undefined (division by zero). Physically, a driven fridge with no work input cannot pump uphill at all — this "infinite COP" is exactly the forbidden second-law case.

Can a Carnot heat pump ever have

COP_{h p} = 1

exactly?

Only in the degenerate limit

T_{C} = 0

K, giving

T_{H} / (T_{H} - 0) = 1

. For any real cold reservoir with

T_{C} > 0

, the Carnot heat pump exceeds 1 — it always drags in some free outside heat.

What is the fridge doing at the instant

T_{C} = T_{H}

(no gap)?

The Carnot COP blows up to infinity, meaning infinitesimal work moves heat. But with no temperature difference there is no reason for heat to have flowed anyway — this is the idealised boundary where "cost" vanishes because the task itself vanishes.

If a real device reports COP equal to the Carnot value, what must be true?

It must be reversible — generating zero entropy, exchanging heat only across infinitesimal temperature differences, and running infinitely slowly. Real machines with finite speed and finite gaps always fall short.

For a heat pump, is it possible for

Q_{C}

(heat pulled from cold) to exceed

Q_{H}

(heat delivered hot)?

No. Since

Q_{H} = Q_{C} + W

with

W \geq 0

, we always have

Q_{H} \geq Q_{C}

. The delivered heat is at least the extracted heat, with equality only when

W = 0

Recall One-line summary of every trap

The recurring lies are: (1) COP is an efficiency (no — it's leverage, often >1); (2) use Celsius (no — kelvin, because $Q / T$ needs an absolute zero); (3) fridge and heat pump COPs are equal (no — they differ by exactly 1); (4) work can be zero or negative (no — $W = Q_{H} - Q_{C} > 0$ ); (5) bigger gap is better (no — the climb $T_{H} - T_{C}$ is the cost).

Connections

Refrigerators and heat pumps — COP — the parent note these traps stress-test.
Heat Engines and Efficiency — contrast COP (leverage) with efficiency (fraction).
Carnot Cycle — source of the reversible ceiling every trap references.
Second Law of Thermodynamics — the law behind "no free uphill heat".
Entropy — the $Q / T$ bookkeeping the sign traps rely on.
First Law of Thermodynamics — the $Q_{H} = Q_{C} + W$ that kills the sign errors.