Question

Assume that liquid water has a temperature-independent capacity of 75 J mol^-1K^-1, and that for the transition below no heat is lost to the surroundings and the processes are reversible.

a. Calculate the chnge in entropy when 1 mol of liquid water is cooled from 100C to 50 C.

b. Calculate the change in entropy of the system when 1 mol of liquid water at 100C is brought into contact with 1 mol of liquid water at 0C.

Answer 1

Let C_p = 75 J mol^-1 K^-1 be the temperature-independent heat capacity of water. Both the processes are reversible.

a) The change in entropy is given by

ΔS = n*C_p*ln T₂/T₁ where n = moles of water taken; T₂ = final temperature = 50⁰C = (50 + 273) K = 323 K and T₁ = 100⁰C = (100 + 273) K = 373 K.

Plug in the values to obtain

ΔS = (1 mol)*(75 J mol^-1K^-1)*ln (323 K/373 K) = -10.794 J K^-1 ≈ -10.8 J K^-1 (ans).

b) Let the common temperature attained by both the bodies of water = T.

Given, T₁ = 100⁰C = 373 K and T₂ = 0⁰C = 273 K

Apply the principle of thermochemistry; heat lost by the hot body = heat gained by the cold body.

Heat lost by water at 100⁰C = n*C_p*(T₁ – T) = (1 mol)*(75 J mol^-1 K^-1)*(373 – T)

Heat gained by water at 0⁰C = n*C_p*(T – T₂) = (1 mol)*(75 J mol^-1 K^-1)*(T – 273)

Equate these two:

(1 mol)*(75 J mol^-1K^-1)*(373 – T) = (1 mol)*((75 J mol^-1K^-1)*(T – 273)

===> 373 – T = T – 273

===> 2T = 373 + 273

===> T = 323

The common temperature attained by both the bodies of water is 323 K.

Now calculate the heat lost by the hot water in the reversible cooling process, Q_rev = (1 mol)*(75 J mol^-1K^-1)*(373 – 323) K = 3750 J.

The change in entropy is ΔS = Q_rev/T = 3750 J/323 K = 11.61 JK^-1 (ans).