91Ó°ÊÓ

The mass of melting ice by engine is\(m = 0.0400\;{\rm{kg}}\)

The time for melting of ice is \(t = 5\;{\rm{min}}\)

The amount of heat required to melt the ice will be equal to the work performed by the engine.

The work performed by the engine to melt ice is given as:

\(W = m\left( {L + c\left( {{T_b} - {T_f}} \right)} \right)\)

Here,\(L\)is the latent heat of ice and its value is\(3.34 \times {10^5}\;{\rm{J}}\),\(c\)is the specific heat of water and its value is\(4200\;{\rm{J}}/{\rm{kg}} \cdot ^\circ {\rm{C}}\),\({T_b}\)and\({T_f}\)are the boiling and freezing temperatures of water and these values are\(100\;^\circ {\rm{C}}\)and\(0\;^\circ {\rm{C}}\).

Substitute all the values in the above equation.

\(\begin{array}{l}W = \left( {0.04\;{\rm{kg}}} \right)\left[ {3.34 \times {{10}^5}\;{\rm{J}} + \left( {4200\;{\rm{J}}/{\rm{kg}} \cdot ^\circ {\rm{C}}} \right)\left( {100\;^\circ {\rm{C}} - 0\;^\circ {\rm{C}}} \right)} \right]\\W = 30160\;{\rm{J}}\end{array}\)

Therefore, the work performed by the engine to melt ice is \(30160\;{\rm{J}}\).