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A mug of coffee cools from l00 °C to room temperature 20 °C. The mass of the coffee is m = 0.25 kg and its specific heat capacity may be assumed to be equal to that of water, c = 4190 J.kg-1 .K -1 . Calculate the change in entropy (a) the coffee (b) the surroundings and (c) the coffee plus the surroundings.

A mug of coffee cools from l00 °C to room temperature 20 °C. The mass of the coffee is m = 0.25 kg and its specific heat capacity may be assumed to be equal to that of water, c = 4190 J.kg-1 .K -1 . Calculate the change in entropy (a) the coffee (b) the surroundings and (c) the coffee plus the surroundings.

Grade:12th pass

1 Answers

Arun
25763 Points
2 years ago
The entropy of both the coffee and surroundings will change, so we have to calculate them separately. For each, we need to find the initial and final states A and B, and calculate the entropy change for a reversible process that takes us between the same states.
Coffee: cools from 𝑇! = 100 °C = 373 K to 𝑇! = 20 °C = 293 K. Consider heat being transferred to the surroundings at an infinitesimal rate
𝑑𝑄 = 𝑚𝑐 𝑑𝑇
! 𝑑𝑄 ! 𝑚𝑐 𝑑𝑇 ! 𝑑𝑇 ∆𝑆= 𝑇= 𝑇 =𝑚𝑐 𝑇
!!!
Then the total entropy change will be
      Hence
i.e. entropy of the coffee decreases, as expected, since heat is flowing out.
∆𝑆coffee =𝑚𝑐 ln𝑇! = 0.25 4190 ln293=−253J.K!! 𝑇! 373
      Surroundings: Heat flow 𝑄 goes into the surroundings, while they remain at 293 K (large heat sink). We calculate 𝑄 by working out how much heat the coffee loses:
𝑄=𝑚𝑐∆𝑇= 0.25 4190 373−293 =8.38×10! J
So the entropy change for the surroundings, at constant temperature 𝑇 = 293 K, is
8.38×10! !! ∆𝑆surroundings = 293 = +286 J. K
which is positive, and larger than the entropy decrease of the coffee. The total entropy
       change is
which is positive, as expected (irreversible process).
∆𝑆tot = ∆𝑆coffee + ∆𝑆surroundings = −253 + 286 = +33 J. K!!

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