Question

a0.5kg piece of metal (c = 600/kgk) at 300 degreee celcius is dumped into a large pool of water at 20 degree celsius. assuming the change in temperature of water to be negligible, caculate the overall change in enthropy for the4 system

Answer 1

Answer on Question #45566 – Physics – Molecular Physics | Thermodynamics

Question:

A 0.5 kg piece of metal (c = 600 J/(kg·K)) at 300 degrees celsius is dumped into a large pool of water at 20 degrees celsius. Assuming the change in temperature of water to be negligible, calculate the overall change in entropy for the system.

Answer:

According to the Second Law of thermodynamics for the reversible processes: $\mathrm{dS} = \frac{\mathrm{dQ}}{\mathrm{T}}$

We assume that piece of metal undergoes an internally reversible heat transfer such that:

$$\mathrm{dS} = \frac{\mathrm{dQ}}{\mathrm{T}} = \frac{\mathrm{m} \cdot \mathrm{c} \cdot \mathrm{dT}}{\mathrm{T}};$$

The assumption that piece of metal has a constant heat capacity allows us to integrate this equation: $\frac{\mathrm{S}_2}{\mathrm{S}_1} \mathrm{dS} = \frac{\mathrm{T}_2}{\mathrm{T}_1} \frac{\mathrm{m} \cdot \mathrm{c} \cdot \mathrm{dT}}{\mathrm{T}}; \quad \Delta S_{\mathrm{Me}} = \mathrm{m} \cdot \mathrm{c} \cdot \ln \mathrm{T} \big|_{\mathrm{T}_1}^{\mathrm{T}_2} = \mathrm{m} \cdot \mathrm{c} \cdot \ln \frac{\mathrm{T}_2}{\mathrm{T}_1}$ ;

This formula uses absolute temperature $T$ in kelvins. We can apply this equation to piece of metal: $\Delta S_{\mathrm{Me}} = 0.5 \, \mathrm{kg} \cdot 600 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \cdot \ln \frac{293 \, \mathrm{K}}{573 \, \mathrm{K}} = -201.2 \, \frac{\mathrm{J}}{\mathrm{K}}$ ;

Assuming the change in temperature of water in the pool to be negligible, we can calculate the change in entropy for it: $\Delta S_{\mathrm{w}} = \frac{\Delta Q}{T_2} = \frac{m \cdot c \cdot \Delta T}{T_2}$

$$\Delta S_{\mathrm{w}} = 0.5 \, \mathrm{kg} \cdot 600 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \cdot \frac{(573 - 293) \, \mathrm{K}}{293 \, \mathrm{K}} = 286.7 \, \frac{\mathrm{J}}{\mathrm{K}};$$

The total change in entropy for the system is equal to the sum of these two entropy changes:

$$\Delta S = \Delta S_{\mathrm{Me}} + \Delta S_{\mathrm{w}} = -201.2 + 286.7 = 85.5 \, \frac{\mathrm{J}}{\mathrm{K}};$$

Answer: $\Delta S = 85.5 \, \frac{\mathrm{J}}{\mathrm{K}}$

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