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## Homework Statement

to illustrate an aspect of the second law of thermodynamics, we regard entropy changes in two cases.

We have a vessel with 1 kg of warm water with T = 80 C and another vessel with 2 kg of cold water with T = 10 C and we mix them.

Case 1:

we pour hot water in the vessel with the cold water. The total entropy change becomes S1

Case 2:

we have a third empty vessel. We mix warm water with 1 kg of the cold water in the vessel. The contents of the third container is then poured with the remainder (1 kg) of cold water. The total entropy change becomes S2

calculate S2 / S1

## Homework Equations

ΔS = m*c*dT/T

ΔS(tot)= ΔS(hot water)+ΔS(cold water)

c=4190 [J/kg K]

T (hot water) = 80 C = 353 K

T (cold water)= 10 C = 283 K

## The Attempt at a Solution

The answer is 1 that means S2 = S1 but I dont get it! I have problem with Case 2

My solution for Case 1 is :

ΔS(hot) = 1 * 4190 * ln(283/353)= -926 J/K

ΔS(cold)=1*4190 * (353-283)/283 = 1036.4 J/K

ΔS1 = 110,4 J/K

Case 2 :

We say that we mix first 1kg warm water with 1 kg cold water:

ΔS (hot) = 1 * 4190 * ln(283/353)= -926 J/K (same !)

ΔS (cold) = (1*4190 * (353-283)/283) = 1036.4 J/K (same !)

Then I will mix the other cold water what I think that when we mixed first 1kg of warm water with cold water, we got a new temperature, when we think of Q (missing) = Q (gain)

when the new value is 45 C, which is 318 K.

Now we have 2 kg of water with temperature 318 K in a third vessel and we want to pour 1 kg of water with temperature 283 K. but when I calculate I get it:

ΔS(remainder cold) = 1 * 4190 * ln(318/283) = 488 J/K !

what is the problem ?! how could I think ?! As I said the solution is S2/S1 = 1 ! But I get a bigger S2