The same reversed cycle sold for warmth. It does not make heat — it moves it, lifting joules from the cold outside and delivering them, plus the work you paid, to the warm inside. So one joule of work can carry several joules of heat past the meter: COPheat = Qh/W = Th/(Th−Tc), always above a resistive heater's flat 1. Down the center the reservoirs go in, the engine computes, the delivered warmth comes out. Blue builds it; red tries to break it.
source W. Thomson (Lord Kelvin), “On the Economy of the Heating or Cooling of Buildings by means of Currents of Air,” Proc. Phil. Soc. Glasgow 3 (1852); repr. Mathematical and Physical Papers — cambridge.org · Art. LX. Rendered, not quoted.
Reverse the engine: pay work W to a compressor, and it drags heat Qc out of the cold reservoir and pushes Qh into the warm one. Energy is conserved: Qh = Qc + W. The heating figure of merit is what you get over what you pay:
COPheat = Qh/W, and for a reversible (Carnot) pump = Th/(Th−Tc). Because Qh = Qc + W, dividing by W gives the exact identity COPheat = COPcool + 1 — the heat pump always out-delivers a resistive heater's COP of 1.
Live figures for the current reservoirs (Kelvin):
Same machine, opposite invoice. A refrigerator is billed for the cold it pulls out — COPcool = Tc/(Th−Tc). A heat pump is billed for the warmth it puts in — COPheat = Th/(Th−Tc). They are the same reversed Carnot cycle, and they differ by exactly one: the work W lands on the warm side too.
Each sphere is the next one's premise: the fridge measures the heat leaving; the pump measures the heat arriving; one added joule of work separates the two counts.
The blue team's live check: recompute the pump over a spread of reservoirs and confirm the laws — COPheat stays > 1, equals COPcool+1, diverges as the gap closes, decays toward 1 as it widens, and always beats resistive. If red tampers, this badge is where it shows.
A heat pump needs two temperatures and a bill. Th is the warm side you deliver to (the room). Tc is the cold source you pull from (the outdoor air or ground). W is the electrical work fed to the compressor, in joules.
All temperatures are absolute (Kelvin). The lift is the gap Th−Tc — the whole cost of the machine is set by how far uphill you push the heat. That is what you feed the panel below.
Move any control — Qh, the COP and the resistive comparison are computed live from Th/(Th−Tc), never looked up.
Left bar: joules of warmth a resistive heater gives for W. Right bar: joules the pump delivers for the same W — the extra height is heat carried in from the cold, free.
What the machine produces, proven: for the current reservoirs the pump turns W joules of work into — joules of warmth at —× — every joule of it obeying Qh = Qc + W and the exact ceiling Th/(Th−Tc). A resistive heater would give exactly W. The pump wins by moving heat instead of dissipating work.
The blue team's witness (left) confirms the laws live; the red team (right) tries to make the count lie.
Worse, the COP collapses exactly when you need heat most: a −15 °C night widens the lift, defrost cycles steal output, and many units fall back to a plain resistive coil (COP 1) below their balance point. The law is real; the ceiling is generous.
“A heat pump is over 100% efficient — it makes heat from nothing.” Cut. It moves heat. Qc is pulled from the environment; only W is bought. COP>1 breaks no conservation law because the extra joules were already out there in the cold.
“COPheat = Tc/(Th−Tc).” Cut. That is the cooling COP. Heating counts the work too, so it is Th/(Th−Tc) — exactly one larger. (This is the red team's tamper below.)
“Heat pumps don't work in the cold.” Kept, corrected. The COP falls as the lift widens, but never below 1 — thermodynamically it always beats resistive heat; only economics and frost, not the law, set the real floor.
The red team's move: swap the heating COP for the cooling formula Tc/(Th−Tc). Now the delivered warmth is understated and, at a wide lift, drops below 1 — the pump appears to lose to a resistive coil. The blue team's witness (window 7) is watching.
Use Tc/(Th−Tc) and the COPheat>1 and COPheat=COPcool+1 checks fail — the witness recomputes, disagrees, and turns red. Nothing is faked; the attack is real and it is caught.