Answer to Question #19386 in Mechanics | Relativity for Joshua

Question

How much heat must a refrigerator remove from 100 g of water at 25 º C to convert it to ice at 0º C? (Specific heat of water = 1 cal /gº C and heat of fusion of ice = 80 cal/g at 0º C)

Answer 1

for cooling down to 0º C refrigerator must remove
100g* (1 cal /gº C) * 25 º C = 2500 cal
for converting into ice refrigerator must remove
100g* 80 cal/g = 8000 cal
Totally 2500+8000 = 10500 cal must be removed