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Cp of ice is 10 J/^oC and that of water is 15 J/^oC, what is the enthalpy change of water from – 15^oC to 0^oC at standard conditions, if the later heat of fusion of water is 110 J?(a) 150 J(b) 315 J(c) 485 J(d) 560 J |
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Answer» The correct choice is (c) 485 J The explanation: ∆H = -15∫^025.dT + 110 = 25*15 + 110 = 485 J. |
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