Question

Cp of ice is 25 J/^oC and that of water is 30 J/^oC, what is the enthalpy change of water from – 10^oC to 35^oC at standard conditions, if the later heat of fusion of water is 110 J?

(a) 250 J

(b) 595 J

(c) 960 J

(d) 1410 J

The question was asked in an internship interview.

The doubt is from Phase Transitions in section Introduction to Energy Balances for Processes without Reaction and Calculation of Enthalpy Changes of Basic Chemical Engineering

Answer 1

The correct answer is (d) 1410 J

To explain I would say: ∆H = -10∫^025.dT + 110 + 0∫^3530.dT = 25*10 + 110 + 35*30 = 1410 J.