Question: A freezer has a coefficient of performance of \(2.40\). The freezer is to convert \(1.80\;kg\) of water at \(25.0^\circ C\) to \(1.80\;kg\) of ice at \( - 5.0^\circ C\) in one hour. (a) What amount of heat must be removed from the water at \(25.0^\circ C\) to convert it to ice at \( - 5.0^\circ C\)? (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?

Coefficient of performance,\(K = 2.40\)

Mass of water, \(m = 180\;{\rm{kg}}\)

Initial temperature of water,\({T_0} = 25^\circ {\rm{C}}\)

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

The expression to calculate the amount of heat when water is converted from \(25^\circ {\rm{C}}\) to \(0^\circ {\rm{C}}\) is given as follows.

\({Q_1} = m{c_w}\Delta T\) …… (i)

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

The expression to calculate the amount of heat when water is converted from \(0^\circ {\rm{C}}\) of water to \(0^\circ {\rm{C}}\) of ice is given as follows.

\({Q_2} = m{L_f}\) …… (ii)

Here,\({L_f}\)is the fusion of water.

The expression to calculate the amount of heat when water is converted from \(0^\circ {\rm{C}}\) of ice to \( - 5^\circ {\rm{C}}\) of ice is given as follows.

\({Q_3} = m{c_{ice}}\Delta T\) …… (iii)

Here,\({c_{ice}}\)is the specific heat of ice.

The expression to calculate the total heatremoved from the water at \(25^\circ {\rm{C}}\) to convert it to ice at \( - 5^\circ {\rm{C}}\) is given as follows.

\({Q_C} = {Q_1} + {Q_2} + {Q_3}\) …… (iv)

The specific heat of the ice,\({c_{ice}} = 2100\;{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\{{\rm{kg}} \cdot {\rm{K}}}}\)

The fusion of water,\({L_f} = 334 \times {10^3}\;{{\rm{J}} \mathord{\left/ {\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\{{\rm{kg}}}}\)

Calculate the amount of heat when water is converted from \(25^\circ {\rm{C}}\) to \(0^\circ {\rm{C}}\).