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Grade: 12th pass
        
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.
10 months ago

Answers : (1)

Arun
22035 Points
							
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!!
10 months ago
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