Mixing hot and cold water Clearly, mixing hot and cold water is an irreversible process. If equal amounts are mixed, the final temperature of the system is the average of the two starting temperatures. Suppose 1.00 kg of water at 100°C is placed in thermal contact with 1.00 kg of water at 0°C.

Question

Mixing hot and cold water

Clearly, mixing hot and cold water is an irreversible process. If equal amounts are mixed, the final temperature of the system is the average of the two starting temperatures. Suppose 1.00 kg of water at 100°C is placed in thermal contact with 1.00 kg of water at 0°C. Eventually the system will equilibrate at 50°C. In this example, we will calculate the approximate change in entropy just after they are mixed, when the hot water has only cooled to 99°C and the cold water has warmed to 1°C. We will assume that the specific heat of water is constant at 4190 J/(kg · K) over the given temperature range.

Accepted Answer

SET UP AND SOLVE This process involves irreversible heat flow because of the temperature differences. The first 4190 J of heat transferred cools the hot water to 99°C (372 K) and warms the cold water from 0°C to 1°C (274 K). The net change in entropy of the hot water is approximately

[latex]\mathrm{\Delta S_{hot}=\frac{-4190 J}{372 K}=-11.3 J/K. } [/latex]

(During this process, the temperature actually varies from 373 K to 372 K, but the resulting error in the calculation is very small.) The change in entropy of the cold water is approximately

[latex]\mathrm{\Delta S_{cold}=\frac{+4190 J}{274 K}=+15.3 J/K. } [/latex]

The total change in entropy of the system during this process is

[latex]\mathrm{\Delta S_{hot}+\Delta S_{cold}=-11.3 J/K+15.3 J/K =+4.0 J/K.} [/latex]

Further increases in entropy occur as the system approaches thermal equilibrium at 50°C (323 K). Because the two temperatures vary continuously during this heat exchange, we would need to use calculus to calculate the entropy changes precisely. It turns out that the total entropy change of the hot water is [latex]\mathrm{\Delta_{Shot} = -603 J/K,}[/latex] the total entropy change of the cold water is [latex]\mathrm{\Delta_{Scold} = 705 J/K,}[/latex] and the total entropy change of the system is 705 J/K - 603 J/K = 102 J/K.

REFLECT The total entropy of the final state is greater than that of the initial state, corresponding to increasing disorder as the distinction between the hotter and colder molecules gradually disappears. We could have reached the same end state by simply mixing the two quantities of water, also an irreversible process. The entropy depends only on the state of the system, and the total entropy change would be the same, 102 J/K.

Practice Problem: Repeat the approximate calculation for the total entropy change when the hot water cools from 76°C to 75°C and the cool water warms from 24°C to 25°C. Answer: 2.0 J/K.

Question 16.8: Mixing hot and cold water Clearly, mixing hot and cold water...

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