A hot piece of copper at 100^{circ} C is drop | vedclass

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(C) The formula for change in entropy is $\Delta S = \int \frac{dQ}{T}$.For the copper,the change in entropy is $\Delta S_{Cu} = \int_{T_i}^{T_f} \fra

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the copper loses entropy and the pond gains more entropy than this amount

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(C) The formula for change in entropy is $\Delta S = \int \frac{dQ}{T}$.For the copper,the change in entropy is $\Delta S_{Cu} = \int_{T_i}^{T_f} \frac{mc dT}{T} = mc \ln(\frac{T_f}{T_i})$,which is negative (the copper loses entropy).For the pond,it has gained heat $Q = mc(T_i - T_f)$. The temperature of the pond $T_{pond} = 30^{\circ} C$ remains constant,so the entropy gained by the pond is $\Delta S_{pond} = \frac{Q}{T_{pond}} = \frac{mc(100 - 30)}{303.15}$.Since this is an irreversible process,the total entropy must increase $(\Delta S_{total} > 0)$.Therefore,$\Delta S_{pond} + \Delta S_{Cu} > 0$,which means $\Delta S_{pond} > |\Delta S_{Cu}|$.Thus,the pond gains more entropy than the entropy lost by the copper.

$A$ hot piece of copper at $100^{\circ} C$ is dropped into a pond at $30^{\circ} C$. The copper cools down to $30^{\circ} C$,while the pond remains at its initial temperature due to its large size. Then,

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