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20 January, 16:57

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?

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  1. 20 January, 17:08
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    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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