b
Heat required to melt $1\,kg$ ice at ${0^ \circ }C$ to water at ${0^ \circ }C$ is
$Q = {m_{ice}}{L_{ice}} = \left( {1\,kg} \right)\left( {80\,cal/g} \right)$
$ = \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$