Question: A refrigerator has a coefficient of performance of \(2.10\). In each cycle it absorbs \(3.10 \times 1{0^4}\;J\) of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

The coefficient of performance,\(K = 2.10\)

The absorbs heat, \({Q_C} = 3.10 \times {10^4}\;{\rm{J}}\)

The coefficient of performance is defined as the ratio of the absorbs heat and work done.

The expression to calculate the coefficient of performance is given as follows.

\(K = \left| {\frac{{{Q_C}}}{W}} \right|\) …… (i)

Here,\({Q_C}\)is the heat absorbs and\(W\)is the work done.

Derive the expression for the work done from the equation (i).

\(W = \frac{{\left| {{Q_C}} \right|}}{K}\)

Substitute \(2.10\) for \(K\) and \(3.10 \times {10^4}\;{\rm{J}}\) for \({Q_C}\) into above equation.

\(\begin{array}{l}W = \frac{{3.10 \times {{10}^4}\;{\rm{J}}}}{{2.10}}\\W = \frac{{310 \times {{10}^4}\;{\rm{J}}}}{{210}}\\W = 1.47 \times {10^4}\;{\rm{J}}\end{array}\)

Hence the mechanical energy is each cycle required to operate the refrigerator is\(1.47 \times {10^4}\;{\rm{J}}\).