Question: A refrigerator has a coefficient of performance of \(2.25\), runs on an input of \(135\;W\)of electrical power, and keeps its inside compartment at \(5^\circ C\). If you put a dozen \(1.0 - L\) plastic bottles of water at \(31^\circ C\) into this refrigerator, how long will it take for them to be cooled down to \(5^\circ C\)? (Ignore any heat that leaves the plastic.)

Coefficient of performance,\(K = 2.25\)

Input power,\(P = 135\;{\rm{W}}\)

Mass of the 1 bottle,\(m = 1\;{\rm{L}}\)

Initial temperature,\({T_0} = 31^\circ {\rm{C}}\)

Final temperature,\({T_1} = 5^\circ {\rm{C}}\)

Number of the bottle put inside the refrigerator is \(12\).

The power is defined as the ratio of the work and time.

The expression to calculate the power is given as follows.

\(P = \frac{W}{t}\) …… (i)

Here,\(W\)is the work and\(t\)is the time.

The expression to calculate the heat when the water bottle is cooling from \(31^\circ {\rm{C}}\) to \(5^\circ {\rm{C}}\) is given as follows.

\({Q_C} = m{c_w}\Delta T\) …… (ii)

Here,\(m\)is the mass,\({c_w}\)is the specific heat of water and\(\Delta T\)is the change in the temperature.

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

\(K = \frac{{{Q_C}}}{W}\) …… (iii)

Hence the time take to cool the bottle inside the refrigerator is \(1.19\;{\rm{h}}\).