Refrigerators and heat pumps — COP: Visual walkthrough

Step 1 — Two reservoirs and the natural downhill flow

WHAT. Draw two big boxes. The top box is hot — call its temperature $T_{H}$ (the little $H$ just means "hot"). The bottom box is cold — temperature $T_{C}$ ( $C$ for "cold"). "Temperature" here is just how hot a thing is, measured on the kelvin scale (we'll see in Step 7 why kelvin and not Celsius).

WHY. Before we pump heat uphill, we must agree what "uphill" means. Heat is energy that moves because of a temperature difference, and on its own it only ever slides from the hot box to the cold box — never the other way. That one-way rule is the whole reason a fridge needs to be plugged in.

PICTURE. The red arrow is the free, natural flow: hot → cold.

Step 2 — Reverse the arrow: this is the machine

WHAT. Now force heat to go the wrong way: pull a chunk $Q_{C}$ out of the cold box and carry it up. The subscript $C$ on $Q_{C}$ means "this is the heat taken from the cold reservoir."

WHY. That is exactly what a refrigerator does: it keeps the inside cold by removing heat from it. But Step 1 said heat won't climb on its own — so something must push. That push is work.

PICTURE. The upward red arrow $Q_{C}$ is now fighting gravity, so to speak. It cannot rise alone.

Step 3 — Where does everything go? The energy balance

WHAT. Follow the energy. Into the machine each cycle: $Q_{C}$ (from the cold box) plus $W$ (from the wall socket). Out of the machine: a chunk dumped into the hot box, call it $Q_{H}$ ("heat to the hot reservoir"). A cycle means the machine's insides return to exactly their starting state, so they store nothing extra.

WHY. Energy cannot appear or vanish (First Law). If the machine keeps nothing, then everything that went in must come out: $in from cold Q_{C} + in from socket W = out to hot Q_{H}$

PICTURE. Two thin arrows go in (bottom + side), one fat arrow comes out on top. Fat = thin + thin.

This matches the First Law of Thermodynamics exactly: $Δ U_{cy c l e} = 0$ .

Step 4 — "How good is it?" is a benefit ÷ cost ratio

WHAT. We want one number that says good or bad. A good fridge moves lots of heat for little work paid. So form the ratio $COP_{r e f} = \frac{what you want}{what you pay} = \frac{Q _{C}}{W} .$ "COP" stands for Coefficient of Performance — just a name for this ratio.

WHY a ratio, and why this one? Because "good" is comparative: 800 J moved is only impressive relative to the work it cost. Dividing benefit by cost gives a leverage number independent of the fridge's size. For the cooling job the benefit is $Q_{C}$ (the heat pulled out of the food).

PICTURE. A balance beam: the thick $Q_{C}$ arrow on the "want" pan, the thin $W$ arrow on the "pay" pan. A big COP tips heavily toward "want."

Step 5 — The heat-pump twist (same machine, different wish)

WHAT. Stand in the room now instead of inside the fridge. You no longer care about $Q_{C}$ ; you care about $Q_{H}$ , the heat delivered into the room. Same machine, new benefit: $COP_{h p} = \frac{Q _{H}}{W} = \frac{Q _{H}}{Q _{H} - Q _{C}} .$

WHY. Nothing about the hardware changed — only which arrow you value. A "heat pump" is a fridge whose fat top arrow you stand under to keep warm.

PICTURE. Same box as Step 3, but a person is drawn in the hot room catching the fat $Q_{H}$ arrow (red); the cold side is faded.

Step 6 — The best possible machine: no wasted entropy

WHAT. So far $Q_{C}$ and $Q_{H}$ could be anything obeying Step 3. Now demand the best machine — a reversible (Carnot) one that wastes nothing. To measure "waste" we use entropy: when heat $Q$ crosses at temperature $T$ , it carries entropy $Q / T$ . Entropy is a bookkeeping quantity that only ever grows for the universe (Second Law); a perfect machine keeps it flat.

WHY this tool — entropy $Q / T$ — and not just energy? Energy balance (Step 3) has infinitely many solutions; it never picks a best one. We need a second equation. The Second Law of Thermodynamics supplies it: the entropy pulled from cold must equal the entropy dumped hot, or the universe's entropy would fall (forbidden) or rise (wasteful).

PICTURE. Two entropy "packets" $Q_{C} / T_{C}$ (leaving cold) and $Q_{H} / T_{H}$ (entering hot), drawn equal in size (the red balance line).

The punchline: the heat ratio is now fixed by temperatures alone, no messy details of the gas.

Step 7 — Substitute: temperatures replace heats

WHAT. Take $COP_{r e f} = \frac{Q _{C}}{Q _{H} - Q _{C}}$ and divide top and bottom by $Q_{C}$ : $COP_{r e f} = \frac{Q _{C} / Q _{C}}{( Q _{H} - Q _{C} ) / Q _{C}} = \frac{1}{\frac{Q _{H}}{Q _{C}} - 1} = \frac{1}{\frac{T _{H}}{T _{C}} - 1} = \frac{T _{C}}{T _{H} - T _{C}} .$

WHY divide by $Q_{C}$ ? Because Step 6 only gave us a ratio $Q_{H} / Q_{C}$ , never the heats separately. Dividing by $Q_{C}$ turns every heat into that known ratio, which we then swap for $T_{H} / T_{C}$ .

PICTURE. A number line of temperature: mark $T_{C}$ and $T_{H}$ ; the gap between them is the "climb" $T_{H} - T_{C}$ . COP is the low temperature over the climb.

Step 8 — Edge cases: what the formula says at the extremes

WHAT & WHY. A formula you trust must behave sanely at its limits.

Tiny gap $T_{H} \to T_{C}$ : denominator $\to 0$ , so $COP \to \infty$ . Pumping heat across no hill costs almost nothing. Makes sense.
Huge gap $T_{C} \to 0$ (or $T_{H}$ far above $T_{C}$ ): numerator small, denominator large, $COP \to 0$ . Pumping heat out of something near absolute zero is nearly impossible — the Second Law bites hardest here.
$W = 0$ with $Q_{C} > 0$ : would need $COP = \infty$ at a finite gap — forbidden. Heat cannot climb for free; you must pay.

PICTURE. COP plotted against the gap $T_{H} - T_{C}$ : a steep curve rocketing to infinity as the gap closes, sinking toward zero as the gap widens (red curve).

The one-picture summary

Everything above, in one frame: heat forced up from cold, work paid in, surplus dumped hot; the two entropy packets balanced (reversible); and the final fraction $T_{C} / (T_{H} - T_{C})$ printed on the temperature gap.

Recall Feynman retelling of the whole walkthrough

Heat is lazy — it only rolls downhill, hot to cold (Step 1). A fridge grabs a scoop of heat from the cold food and shoves it uphill into the warm room, which it can only do by paying with a push of electricity (Steps 2–3). "How good is it?" is just heat carried up ÷ push spent — a big number is good (Step 4). Stand in the room instead of the fridge and you value the fat top arrow, which always contains your push too, so a heat pump scores exactly one point higher (Step 5). The best possible machine wastes no entropy: the entropy it lifts from cold equals the entropy it drops hot, and that single balance nails the heat ratio to $T_{H} / T_{C}$ (Step 6). Swap that in and the messy heats vanish, leaving a clean fraction: cold temperature over the climb (Step 7). Test it at the edges — no hill means free lifting (COP → ∞), a giant hill means near-impossible lifting (COP → 0) — and it all makes sense (Step 8). Just remember: kelvin, always kelvin.

Connections

First Law of Thermodynamics — the energy balance of Step 3.
Second Law of Thermodynamics — why the work must be paid, and the entropy balance of Step 6.
Entropy — the $Q / T$ bookkeeping that gives the reversible heat ratio.
Carnot Cycle — the reversible machine that reaches the ceiling.
Heat Engines and Efficiency — the same $Q_{H}, Q_{C}, W$ run forwards.