Move heat from cold to warm with work

Reverse the service, not just a sign

A refrigerator’s useful service is removing heat from a cold region. A heat pump’s heating service is delivering heat to a warmer region. Both require a full energy account. In this lab, a proposed cyclic device removes Qc = 1000 J from a cold reservoir and receives work Win. It delivers Qh to a reservoir fixed at 300 K.

Qc, Qh and Win are positive magnitudes. The retained working substance returns to its starting state, so its internal-energy change is zero. Energy conservation gives Qh = Qc + Win. This differs from a heat engine delivering net work: here work enters the device.

At the default Win = 300 J, the warm side must receive 1300 J. It receives both the 1000 J removed from the cold side and the 300 J work input. If you use the earlier convention that work by the device is positive, its work is −300 J and its net heat is 1000−1300 = −300 J; ΔU = Q−W is still zero.

Check entropy before accepting a proposal

The cold reservoir is initially set to Tc = 240 K and stays at that temperature by assumption. It loses 1000 J, so its entropy change is −1000/240 ≈ −4.166667 J/K. The 300 K warm reservoir gains 1300 J, giving +4.333333 J/K. The device returns to its original entropy, and ideal mechanical work storage has no entropy change. The total is +0.166667 J/K.

This passes a necessary thermodynamic check. It does not identify a refrigerant, compressor, electrical motor or practical operating speed. The selected cold temperature is a teaching scenario, not a suggested household thermostat setting.

Reduce work input to 200 J while holding 240 K. Energy still balances: 1000 + 200 = 1200 J delivered. But the warm reservoir gains only 1200/300 = 4 J/K, making the combined change −0.166667 J/K. The proposal is rejected. Displaying its calculated heat transfer does not mean the device can perform it.

Find the minimum work

The combined entropy condition is Qh/Th − Qc/Tc ≥ 0, where Th = 300 K. Substitute Qh = Qc + Win and rearrange:

Win ≥ Qc(Th − Tc)/Tc.

The temperatures are in kelvin; their ratio uses an absolute scale. For our default reservoirs, Win must be at least 1000 × (300−240)/240 = 250 J. Select 250 J and verify that the reservoir entropy changes cancel. This is an ideal reversible boundary. Ordinary irreversible operation needs more work for the same heat removal and reservoir temperatures.

Set work to zero. The proposed removal remains 1000 J, but the total entropy change becomes 1000/300−1000/240 ≈ −0.833333 J/K. The proposal is impossible under these assumptions. A performance ratio that divides by zero is undefined; it is not an infinitely effective refrigerator.

Connect the balance to a device concept

A vapor-compression refrigerator uses a circulating working fluid. In a simplified cycle, the fluid absorbs heat on the cold side, compression requires work, heat is released on the warm side, and a pressure-reduction stage prepares the fluid to absorb heat again. The OpenStax refrigerator reference provides background. Our lab accounts for net transfers; it does not model phase changes, refrigerant properties or component losses.

With a refrigerator door open inside a closed room and its heat exchanger in that same room, both heat flows act on the room. Their difference adds the work input as heat over complete cycles. An air conditioner must deliver heat outside the region it is intended to cool. Always draw the boundary before deciding whether a region is cooled.

Transfer to another cooling target

A proposed refrigerator removes 600 J per cycle from 250 K and delivers heat to 300 K. Find the minimum reversible work. Screen a 100 J work proposal and calculate its warm-side heat delivery.

Check the refrigeration transfer

Minimum work is 600(300−250)/250 = 120 J. At 100 J input, warm-side delivery would be 700 J. The total entropy change would be 700/300−600/250 ≈ −0.066667 J/K. The claim balances energy but fails the second-law screen.