Find the change in the entropy in the following process $100 \,gm$ of ice at $0°C$ melts when dropped in a bucket of water at $50°C$ (Assume temperature of water does not change) ..... $ cal/K$
Diffcult
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(b)Gain of entropy of ice
${S_1} = \frac{{\Delta Q}}{T} = \frac{{mL}}{T} = \frac{{80 \times 100}}{{(0 + 273)}} = \frac{{8 \times {{10}^3}}}{{273}}cal/K$
Loss of entropy of water $ = {S_2} = - \frac{{\Delta Q}}{T} = - \frac{{mL}}{T}$
$ = \frac{{80 \times 100}}{{(273 + 50)}} = \frac{{8 \times {{10}^3}}}{{323}}cal/K$
Total change of entropy
${S_1} + {S_2} = \frac{{8 \times {{10}^3}}}{{273}} - \frac{{8 \times {{10}^3}}}{{323}} = + 4.5\;cal/K$
${S_1} = \frac{{\Delta Q}}{T} = \frac{{mL}}{T} = \frac{{80 \times 100}}{{(0 + 273)}} = \frac{{8 \times {{10}^3}}}{{273}}cal/K$
Loss of entropy of water $ = {S_2} = - \frac{{\Delta Q}}{T} = - \frac{{mL}}{T}$
$ = \frac{{80 \times 100}}{{(273 + 50)}} = \frac{{8 \times {{10}^3}}}{{323}}cal/K$
Total change of entropy
${S_1} + {S_2} = \frac{{8 \times {{10}^3}}}{{273}} - \frac{{8 \times {{10}^3}}}{{323}} = + 4.5\;cal/K$
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