A 5-kg block initially at 623 K is quenched in an insulated tank that contains 1

Question

A 5-kg block initially at 623 K is quenched in an insulated tank that contains 100 kg of water at 303 K. Assume the block and the water to be incompressible substances (cblock = 0.45 kJ/(kg.K) and cwater = 4.18 kJ/(kg.K)).
Determine:
a) (5 points) the final equilibrium temperature inside the insulated tank, Tf;
b) (10 points) the change in the entropy of the block and of the water separately.
c) (10 points) the total created entropy during the quenching process of the combined system (content of insulated tank + surroundings). How does this relate to the second law of thermodynamics?

Explanation / Answer

Apply conservation of energy to our block and water system. And heat equation mcT

heat lost by block = heat gained by water

mblock cblock (623-T) = mwater cwater (T - 303)

Solve for T by putting in the numbers

5 * 4500 (623 -T) = 100 * 4180 (T - 303)

22500 (623 - T) = 418000 (T - 303)

14017500 - 22500T = 418000T - 126654000

440500T = 140671500

T = 319K

b) change in entropy of the block is S = q/T = 623-319 / 319 = 0.95 J/K-1

change in entropy of water is S = q/T = 319 - 303 / 319 = 0.05J/K-1

c)The total change in entropy is S1 + S2 = 1J/K-1

This relates that the total entropy of the universe is always increasing in any process or system.

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