Q3(b). Entropy changes when an ice cube melts in the sea at 0 °C
A 74.6 g ice cube floats in the sea. The temperature and pressure of the system and surroundings are 0 °C and 1 atm. Calculate ΔSsyst, ΔSsurr and ΔSuniv for the melting of ice cube. What can you conclude about the nature of process from the value of ΔSuniv? (The molar heat of fusion of water is 6.01 kJ/mol)
Answer in brief
ΔSsyst = +91.1 J K⁻¹, ΔSsurr = −91.1 J K⁻¹, ΔSuniv = 0; the melting is reversible: ice and water are in equilibrium at 0 °C and 1 atm.
Part of VKC Basic
Read the key points and the full model answer with VKC Basic
VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversentropy of fusionsecond law of thermodynamicsentropy of universereversible processphase equilibrium
All parts of Question 3
- 3(a) The least stable of OCN⁻, ONC⁻ and SCN⁻10 marks
- 3(b) Entropy changes when an ice cube melts in the sea at 0 °C10 marks
- 3(c) Speed distribution curves of argon and helium10 marks
- 3(d) Phase diagram of water; critical point and triple point10 marks
- 3(e) Forces on an ion moving through a strong-electrolyte solution10 marks