Water's heat of fusion is 80 cal/g. its specific heat is 1.0 cal/g degrees celsius, and its heat of vaporization is 540 cal/g. a canister is filled with 330g of ice and 100g of liquid water, both at 0 degrees celsius. the canister is placed in an oven until all the water has boiled off and the canister is empty. how much energy in calories was absorbed?

First, write out the process of heat absorption: 
 Ice --(heat of fusion)--> Water at 0 degreesC --(specific heat)--> Water at 100 degrees C --(heat of vaporization)--> steam 
 (330 g ice)(80 cal/g) = 26,400 cal
 Now there are 430 g of liquid water at 0 degrees C. 
 (430 g liquid water)(1 cal/gC)(100 C - 0 C) = 43,000 cal
 (430 g liquid water)(540 cal/g) = 232,200 cal 
 In total, 
 26,400 + 43,000 + 232,200 = 3 x 10^5 cal

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