Question

Suppose 1 mole of water is initially at T=253K and P=1atm. The temperature is
raised to T=400K by placing it in contact with hot iron pan that is maintained at
T=400K. The pressure is maintained at P=1atm. Assume the heat capacity at constant
pressure of ice CP is constant between 253K and T=273K and equals 37 J K
-1mol-1. Assume the heat capacity CP of liquid water is constant and is equal to 75.3 J K
-1mol-1. Assume finally that CP of water vapor in the temperature range given is 33 JK
-1mol-1. For water ∆Hmelt=6010Jmol-1and ∆Hvap=40100 Jmol-1. Enthalpy change of water and iron pan?

Answer 1

ok, take the temperature range of the solid ice so 273K-253K= 20K
this range of temperature you multiply with your ice cp which is 37J/K.
do that for the ranges with liquid and vapor, so 273-373 for liquid and 373 to 400 for vapor. add all these values. then add the changes in enthalpy in the transition state from solid to liquid and then liquid to vapor. This total Energy is the change in enthalpy of the water and iron pan. So that energy is going from the iron pan to the water.
hope that helps :)