It takes approximately 27.78 J of work per cycle to move the heat from the 5.00 °C interior to the 20.0 °C exterior region in the refrigerator.
The coefficient of performance (COP) of a refrigerator is defined as the ratio of the amount of heat it absorbs from the interior to the work done on it. In this case, the COP is given as 4.5.
To find the amount of work done per cycle to move the heat from the interior (5.00 °C) to the exterior (20.0 °C) region, we can use the formula:
COP = Q_in / W
where COP is the coefficient of performance, Q_in is the heat absorbed, and W is the work done.
Given that the COP is 4.5 and the heat absorbed (Q_in) is 125 J, we can rearrange the formula to solve for W:
W = Q_in / COP
Substituting the values, we get:
W = 125 J / 4.5
Calculating this, we find that the work done per cycle is approximately 27.78 J.