Question

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

Answer 1

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

**Solution:**

We assume that piece of metal undergoes an internally reversible heat transfer such that $dS = \frac{dQ}{T} = \frac{mcdT}{T}$. The assumption that piece of metal has a constant heat capacity allows us to integrate this equation giving $\Box \Delta S = m \ln \left(\frac{T2}{T1}\right)$. In this calculation the temperature must be in kelvins. We can apply this equation to piece of metal, here using units of kelvins for the heat capacity.

$$\Delta S_{1} = 0.5 \mathrm{kg} * 600 \frac{\mathrm{J}}{\mathrm{kg} \mathrm{K}} * \ln \frac{293 \mathrm{K}}{573 \mathrm{K}} = -201,24 \frac{\mathrm{J}}{\mathrm{K}}$$

Assuming the change in temperature of water to be negligible, we can calculate change in entropy for the large pool of water

$$\begin{array}{l} dS = \frac{dQ}{T} \rightarrow \Delta S_{2} = \Box \frac{Q}{T} = \frac{m c \Box T}{T2}, \\ \Delta S_{2} = 0.5 \mathrm{kg} * 600 \frac{\mathrm{J}}{\mathrm{kg} \mathrm{K}} * \frac{(300 - 20) \mathrm{K}}{293 \mathrm{K}} = 286,68 \frac{\mathrm{J}}{\mathrm{K}} \\ \end{array}$$

where $\Delta \Box Q$ is the amount of heat received from the piece of metal

The overall change in entropy for the system is simply the sum of these two entropy changes.

$$\Delta S = \Delta S_{1} + \Delta S_{2} = -201,24 \frac{\mathrm{J}}{\mathrm{K}} + 286,68 \frac{\mathrm{J}}{\mathrm{K}} = 85,44 \frac{\mathrm{J}}{\mathrm{K}}.$$

Answer: $\Delta S = 85,44 \frac{\mathrm{J}}{\mathrm{K}}$.