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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?

Respuesta :

W0lf93
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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