Copper immersed in water: total ΔS

Problem

A copper block of mass $m=0{,}2$ kg (specific heat $c_s = 385\ \text{J/(kg}\cdot\text{K)}$) at temperature $T_A = 350$ K is immersed in a large water reservoir at $T_\text{atm} = 290$ K, until thermal equilibrium is reached. Calculate the entropy changes of the copper, of the water, and the total. Is the process reversible?

Solution

The reservoir is large, so it behaves as a thermostat: its temperature stays at $T_\text{atm}=290$ K. The copper cools from $350$ K to $290$ K.

Entropy of the copper (variable temperature, integrating $\dd Q = m c_s\,\dd T$): $\Delta S_\text{copper} = m c_s\ln\frac{T_\text{atm}}{T_A} = 0{,}2\cdot 385\cdot\ln\frac{290}{350} = -14{,}5\;\text{J/K}$

Heat released by the copper and absorbed by the water: $Q = m c_s (T_A - T_\text{atm}) = 0{,}2\cdot 385\cdot 60 = 4620\;\text{J}$

The water receives this heat at the constant temperature $T_\text{atm}$: $\Delta S_\text{water} = \frac{Q}{T_\text{atm}} = \frac{4620}{290} = +15{,}9\;\text{J/K}$

Total balance: $\Delta S_\text{tot} = \Delta S_\text{copper} + \Delta S_\text{water} = -14{,}5 + 15{,}9 = +1{,}45\;\text{J/K}$ $\ev{\Delta S_\text{tot} \approx 1{,}45\;\text{J/K}}$ Since $\Delta S_\text{tot} > 0$, the process is IRREVERSIBLE: heat flows spontaneously across a finite temperature difference.

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Secondo principio della termodinamica · Calore specifico e capacità termica Skills: Equazione calorimetrica · Bilancio entropico Objects: Calorimetro