Blundell Thermal Physics · Problem 14.1

Entropy change during cooling involves integrating $dS = \frac{dQ}{T}$ over the process—since heat flows out of the tea, $dQ$ is negative, making the entropy change negative too. The key insight is that entropy of the tea alone decreases because it's an open system losing thermal energy to a warmer surroundings; however, the total entropy of the universe (tea plus environment) increases, consistent with the second law. This apparent paradox resolves when you account for the much larger entropy gain of the cooler environment absorbing that heat.

Entropy change during cooling involves integrating $dS = \frac{dQ}{T}$ over the process—since heat flows out of the tea, $dQ$ is negative, making the entropy change negative too. The key insight is that entropy of the tea alone decreases because it's an open system losing thermal energy to a warmer surroundings; however, the total entropy of the universe (tea plus environment) increases, consistent with the second law. This apparent paradox resolves when you account for the much larger entropy gain of the cooler environment absorbing that heat.