Coefficient of Performance, COP:

$\overline{){\mathbf{COP}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{Q}}_{\mathbf{C}}}{\mathbf{W}}{\mathbf{=}}\frac{{\mathbf{Q}}_{\mathbf{C}}}{{\mathbf{Q}}_{\mathbf{H}}\mathbf{-}{\mathbf{Q}}_{\mathbf{C}}}{\mathbf{=}}\frac{{\mathbf{T}}_{\mathbf{C}}}{{\mathbf{T}}_{\mathbf{H}}\mathbf{-}{\mathbf{T}}_{\mathbf{C}}}}$

Q_{C} is the heat removed from inside and W is the work done. The two must be equal for the coefficient of performance to be 1.

Q_{H} is the heat supplied by the hot reservoir/outside. T_{C} is the temperature of the cold body and T_{H} is the temperature of the hot body.

How would you increase the coefficient of performance of an ideal refrigerator?

a) Increase the mechanical work input.

b) Decrease the outside temperature.

c) Decrease the inside temperature.

d) Increase the outside temperature.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Entropy and the Second Law concept. You can view video lessons to learn Entropy and the Second Law. Or if you need more Entropy and the Second Law practice, you can also practice Entropy and the Second Law practice problems.