Compare cooling and heating performance

Choose what service you are counting

At the default proposal, the device removes 1000 J from the cold reservoir, receives 300 J of work and delivers 1300 J to the warm reservoir. A cooling user values the 1000 J removed. A heating user values the 1300 J delivered. The work input is the resource used in both comparisons.

The coefficient of performance, or COP, is the useful heat service divided by work input. For cooling, COPc = Qc/Win = 1000/300 ≈ 3.333. For heating, COPh = Qh/Win = 1300/300 ≈ 4.333. These ratios have no units and are not percentages of work converted into heat. The device transfers heat already present in the cold reservoir, in addition to receiving work.

For this same cycle and boundary, Qh = Qc + Win. Dividing by positive Win gives COPh = COPc + 1. Check that relation at several settings. It need not describe two unrelated appliances with different conditions or accounting boundaries. At zero work input the ratios are undefined, and the finite-transfer proposal in this lab is rejected.

Derive bounds for the two services

The second law gives Win ≥ Qc(Th−Tc)/Tc. Dividing the cooling service Qc by positive work yields the reversible bound:

COPc ≤ Tc/(Th−Tc).

Adding one gives COPh ≤ Th/(Th−Tc). Both use absolute temperatures with Th > Tc > 0. They describe a device exchanging heat only with these two reservoirs and receiving work, with no other changes. The reversible equality case follows from the same thermal reasoning as the Carnot-cycle reference.

At Tc = 240 K and Th = 300 K, cooling COP cannot exceed 4, and heating COP cannot exceed 5. The default 300 J proposal is below both limits and passes the entropy screen. At 250 J it reaches both limits. At 200 J it proposes cooling COP 5 and heating COP 6, exceeding the limits and failing the screen. A larger reported COP is not automatically a credible improvement.

Change the temperature lift

Keep work input at 300 J while changing the cold reservoir temperature. Heat removal stays 1000 J and warm-side delivery stays 1300 J, so the proposed COPs stay 3.333 and 4.333. However, the permissible limits change.

At 200 K, the temperature lift to 300 K is 100 K. Minimum work is 500 J; maximum cooling COP is 2 and heating COP is 3. The 300 J proposal is rejected, with total entropy change 1300/300−1000/200 = −0.666667 J/K.

At 280 K, the lift is only 20 K. Minimum work is about 71.429 J, cooling COP is bounded by 14 and heating COP by 15. The same 300 J proposal passes the entropy screen, with combined change about +0.761905 J/K. A smaller temperature lift relaxes this ideal work requirement; it does not guarantee that a particular real machine achieves the bound.

Separate a ratio from capacity

The lab uses energy per cycle and specifies no cycle rate. COP alone does not tell you how fast a room warms or a freezer removes heat. If a separate assumption supplied two cycles per second for the default proposal, the cooling rate would be 2000 J/s = 2 kW, the work-input rate 0.6 kW, and the heating rate 2.6 kW. These rates are conditional arithmetic, not a performance prediction from this model.

A measured system-level COP would need a stated operating period, temperatures and boundary, including the relevant compressor, fan, pump and control inputs. Seasonal performance needs evidence across changing conditions; one ideal cycle is insufficient. The work input here is mechanical, so no electricity use or cost is inferred without conversion and operating information.

Transfer using a heating specification

A heat pump delivers 4500 J per cycle to the warm side while receiving 1500 J of work. Calculate the heat removed from the cold side, both COPs, and explain why the heating ratio does not violate energy conservation. Would the energy figures alone prove it can operate between any two reservoir temperatures?

Check the heat-pump transfer

Cold-side removal is 4500−1500 = 3000 J. Cooling COP is 3000/1500 = 2; heating COP is 4500/1500 = 3. The extra delivered heat comes from the cold side, so energy balances. Feasibility still depends on reservoir temperatures and the entropy balance; energy figures alone do not establish it.