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The heat input to the heat engine is the sum of the work performed by the engine and the heat rejected from the engine.

The inside temperature of the room is\({T_{\rm{L}}} = 21{\rm{^\circ C}}\).

The outside temperature of the room is \({T_{\rm{H}}} = 32{\rm{^\circ C}}\).

The expression for the heat input to the system is

\({Q_{\rm{H}}} = W + {Q_{\rm{L}}}\).

Here, \(W\) is the work done and \({Q_{\rm{L}}}\) is heat output from the system.

The expression for the efficiency of the heat engine is

\(\begin{aligned}{c}e &= \frac{W}{{{Q_{\rm{H}}}}}\\e &= \frac{W}{{W + {Q_{\rm{L}}}}}.\end{aligned}\) … (i)

The expression for the efficiency of the heat engine in terms of temperature is

\(e = 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\). … (ii)

Equate equations (i) and (ii).

\(\begin{aligned}{c}\frac{W}{{W + {Q_{\rm{L}}}}} &= 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\\W &= {Q_{\rm{L}}}\left( {\frac{{{T_{\rm{H}}}}}{{{T_{\rm{L}}}}} - 1} \right)\\\frac{W}{t} &= \frac{{{Q_{\rm{L}}}}}{t}\left( {\frac{{{T_{\rm{H}}}}}{{{T_{\rm{L}}}}} - 1} \right)\end{aligned}\)

The electric power required for \(4.8\;{\rm{kW}}\) radiation is calculated below:

\(\begin{aligned}{l}{\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} &= \left( {4.8\;{\rm{kW}}} \right)\left( {\frac{{{{10}^3}\;{\rm{W}}}}{{1\;{\rm{kW}}}}} \right)\left( {\left( {\frac{{\left\{ {\left( {32{\rm{^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}{{\left\{ {\left( {{\rm{21^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}} \right) - 1} \right)\\{\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} &= 179.59\;{\rm{W}}\end{aligned}\)

The electric power required for \(500\;{\rm{W}}\) radiation is calculated below:

\(\begin{aligned}{l}{\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}} &= \left( {500\;{\rm{W}}} \right)\left( {\left( {\frac{{\left\{ {\left( {32{\rm{^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}{{\left\{ {\left( {{\rm{21^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}} \right) - 1} \right)\\{\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}} &= 18.7\;{\rm{W}}\end{aligned}\)

The amount of electric power that would be saved is calculated below:

\(\begin{aligned}{l}W &= {\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} - {\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}}\\W &= \left( {179.59\;{\rm{W}}} \right) - \left( {18.7\;{\rm{W}}} \right)\\W &= 161\;{\rm{W}}\end{aligned}\)

Thus, the amount of electric power that would be saved is \(161\;{\rm{W}}\).