બરફની ગુપ્ત ઉષ્મા $80\, cal/gm$ લો.
\(Q = {m_{ice}}{L_{ice}} = \left( {1\,kg} \right)\left( {80\,cal/g} \right)\)
\( = \left( {1000\,g} \right)\left( {80\,cal/g} \right) = 8 \times {10^4}\,cal\)
\(Change\,in\,entropy,\,\Delta S = \frac{Q}{T} = \frac{{8 \times {{10}^4}\,cal}}{{\left( {273\,K} \right)}}\)
\( = 293\,cal/K\)