Discussion
A 5 kg metal block (\(c_{p}\) = 0.5 kJ/kg-K) at 373 K is submerged into 10 kg of water (\(c_{p}\) = 4.2 kJ/kg-K) at 293 K in an insulated rigid container without spilling. Assuming thermal equilibrium is reached, the approximate entropy change of the universe is
Answer: B
The entropy change of the universe is about 0.073 kJ/K.
The container is insulated and rigid, so the heat lost by the block equals the heat gained by the water:
\[ \begin{aligned} 5(0.5)(373 - T) &= 10(4.2)(T - 293) \\ T &= \frac{932.5 + 12306}{44.5} = 297.5\ \text{K} \end{aligned} \]
Each body's entropy change is \(mc\ln(T_{2}/T_{1})\):
\[ \begin{aligned} \Delta S_{block} &= 2.5\ln\tfrac{297.5}{373} = -0.565 \\ \Delta S_{water} &= 42\ln\tfrac{297.5}{293} = +0.639 \\ \Delta S_{univ} &\approx 0.073\ \text{kJ/K} \end{aligned} \]
No heat leaves the container, so the surroundings' entropy does not change. \(-0.565\) kJ/K is the block's change alone, and 0.642 kJ/K is close to the water's alone; the universe's total must be positive for this irreversible process.
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