When $1\, kg$ of ice at $0^o C$ melts to water at $0^o C,$ the resulting change in its entropy, taking latent heat of ice to be $80\, cal/gm,$ is ...... $cal/K$
AIPMT 2011, Medium
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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$
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