Question

200 g of hot gold at 120.0^oC (c_s= 0. 131 J g⁻¹ K⁻¹) is placed in a thermally insulated container that has 25.0 g of water in it at 10.0^oC (c_s= 4. 186 J g⁻¹ K⁻¹).

a) What is ∆S_Au? [-6.64 J K⁻¹]
b) What is ∆S_H2O?

Now carefully consider what the system is and what the surroundings are and answer these questions.
c) Is the process reversible?
d) Calculate the value of ∆S_uni and show that it confirms your answer to part c).

Answer 1

Let the mixture finally settles to a constant temperature T.

Heat lost by gold = Heat gained by water

ms∆T = ms∆T

200*0.131*(120-T) = 25*4.186*(T-25)

3144 - 26.2T = 104.65T - 2616.25

130.85T = 5760.25

T = 44^oC

Q_Au = -200*0.131*(120-44) = -1991.2 J

Q_H2O = 25*4.186*(44-25) = 1988.35 J

T = 44^oC = 317 K

a. ∆S_Au​ = ​-Q_Au/T = 1991.2/317

∆S_Au​ = 6.28 J/K

b. ∆S_H2O = -Q_H2O/T = -1988.35/317

∆S_H2O = -6.27 J/K

c. No, the process is not reversible.

d. ∆S_uni = ∆S_Au​ + ∆S_H2O

∆S_uni = 6.28 - 6.27 = 0.01 J/K

Since, ∆S_uni ≠ O. The process is not reversible.