Posted by kat on Sunday, February 1, 2009 at 9:00pm.
From the information given, I think it is the enthalpy, ∆H, you need to find for the overall process.
You have the following calculations:
1. Heating of ice from 240 to 273 K:
∆H1 = (36 j/mol.ēK)(4.0 mol)(273K-240K)
2. Melting of ice:
∆H2 = (4.0mol.)(5.0kJ/mol
3. Heating of water:
∆H3 = (74 J/mol.ēK)(4.0mol)(310ēK-273ēK)
Calculate and add the 3 enthalpies and add them up.
NOTE: 310 K is below the boiling of water. No significant evaporation takes place.
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