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Heat of fusion of benzene is 10 J at 0^oC, if Cp of solid benzene is 10 J/^oC, and that of liquid benzene is 5 J/^oC what is the enthalpy change of benzene from – 10^oC to 10^oC?

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Heat of fusion of benzene is 10 J at 0^oC, if Cp of solid benzene is 10 J/^oC, and that of liquid benzene is 5 J/^oC what is the enthalpy change of benzene from – 10^oC to 10^oC?

(a) 160 J

(b) 200 J

(c) 240 J

(d) 300 J

The question was posed to me in homework.

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

1 Answer

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Correct option is (a) 160 J

Best explanation: ∆H = -10∫^010.dT + 10 + 0∫^105.dT = 10*10 + 10 + 5*10 = 160 J.

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