1.

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

Answer» 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.


Discussion

No Comment Found

Related InterviewSolutions