Question

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is :

(Specific heat of water liquid and water vapour are 4.2 kJ K^-1 kg^-1 and 2.0 kJ K^-1 kg^-1; heat of liquid fusion and vapourisation of water are 334 kJ^-1 and 2491 kJ kg^-1, respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)

• A. 7.90 kJ kg^-1 K^-1
• B. 2.64 kJ kg^-1 K^-1
• C. 8.49 kJ kg^-1 K^-1
• D. 9.26 kJ kg^-1 K^-1

Hint:

Add four entropy terms: melting at 273 K, heating liquid 273 K to 373 K, vaporisation at 373 K, and heating vapour 373 K to 383 K.

Answer 1

The Correct Option is D

Take 1 kg of water. Total entropy change = sum of four steps.
(i) Fusion at 273 K: \(\Delta S_1 = \frac{334}{273} = 1.223\) kJ kg^-1 K^-1.
(ii) Heating liquid from 273 K to 373 K: \(\Delta S_2 = 4.2\ln\frac{373}{273} = 4.2 \times 2.303 \times (2.572-2.436) = 4.2 \times 2.303 \times 0.136 = 1.315\).
(iii) Vaporisation at 373 K: \(\Delta S_3 = \frac{2491}{373} = 6.678\).
(iv) Heating vapour from 373 K to 383 K: \(\Delta S_4 = 2.0\ln\frac{383}{373} = 2.0 \times 2.303 \times (2.583-2.572) = 2.0 \times 2.303 \times 0.011 = 0.051\).
Total = 1.223 + 1.315 + 6.678 + 0.051 = 9.27, i.e. about 9.26 kJ kg^-1 K^-1.
Answer: (4) 9.26 kJ kg^-1 K^-1