Find the change in the entropy in the followi | vedclass

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(B) The process involves two parts: the melting of ice and the heat loss from the water.$1$. Entropy change of ice $({\Delta S}_{ice})$: The ice melts

Vedclass · Accepted Answer

$+4.5$

Vedclass · Answer

(B) The process involves two parts: the melting of ice and the heat loss from the water.$1$. Entropy change of ice $({\Delta S}_{ice})$: The ice melts at $T = 0^{\circ}C = 273 \, K$. The heat absorbed is $Q = mL = 100 \, g 	imes 80 \, cal/g = 8000 \, cal$.${\Delta S}_{ice} = \frac{Q}{T} = \frac{8000}{273} \approx 29.30 \, cal/K$.$2$. Entropy change of water $({\Delta S}_{water})$: The water loses the same amount of heat at $T = 50^{\circ}C = 323 \, K$.${\Delta S}_{water} = -\frac{Q}{T} = -\frac{8000}{323} \approx -24.77 \, cal/K$.$3$. Total entropy change $({\Delta S}_{total})$: ${\Delta S}_{total} = {\Delta S}_{ice} + {\Delta S}_{water} = 29.30 - 24.77 = +4.53 \, cal/K \approx +4.5 \, cal/K$.

Find the change in the entropy in the following process: $100 \, g$ of ice at $0^{\circ}C$ melts when dropped in a bucket of water at $50^{\circ}C$. (Assume the temperature of the water does not change). The result is in $cal/K$.

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