Heat flow for a fridge or air conditioner

This simulation shows the energy flow for a fridge or air conditioner. It takes some amount of work (W) to remove 100 J (QL) from a colder temperature location, with the total of the W and QL being the net heat (QH) that is dumped into the higher temperature region.

The minimum amount of work needed to remove a certain amount of heat from the colder location depends on the two temperatures the device is operating between. For a given pair of temperatures, the ideal coefficient of performance (COP) of the device is given by:
ideal COP = TL/(TH - TL) = QL/Wmin

Real devices are less efficient than the ideal case. The slider at the bottom allows you to adjust the coefficient of performance to be more realistic than the ideal case.

Simulation written by Andrew Duffy, and first posted on 12-01-2018.

Creative Commons License
This work by Andrew Duffy is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
This simulation can be found in the collection at http://physics.bu.edu/~duffy/classroom.html.

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