ΔS in two heat exchanges

Problem

A hot block at $400\ \text{K}$ is brought into contact with a cold one at $300\ \text{K}$. Calculate the entropy change of each and the total change, assuming $100\ \text{J}$ passes from the hot to the cold block. Discuss the sign.

Solution

We treat the two blocks as fixed-temperature reservoirs (the exchange is small compared with their heat capacity), so $\Delta S=\dfrac{Q}{T}$ with the sign of the heat. Hot block, gives up $100\ \text{J}$ at $400\ \text{K}$: $\Delta S_c=\frac{-100}{400}=-0{,}25\;\text{J/K}$ Cold block, receives $100\ \text{J}$ at $300\ \text{K}$: $\Delta S_f=\frac{+100}{300}\approx +0{,}333\;\text{J/K}$ Total: $\Delta S_\text{tot}=\Delta S_c+\Delta S_f\approx -0{,}25+0{,}333=+0{,}083\;\text{J/K}>0$ The hot block loses little entropy (it gives it up at a high temperature) compared with how much the cold block gains (it receives it at a low temperature): the net balance is positive, the process is irreversible. $\ev{\Delta S_\text{tot}\approx +0{,}083\;\text{J/K}}$

Links

Topics: Entropy and the second law Concepts: Entropy · Second law of thermodynamics Skills: Entropy balance