the initial temperature of 150 g of ice is -20 degree C. The specific heat capac

Question

the initial temperature of 150 g of ice is -20 degree C. The specific heat capacity of ice is 0.5 cal/g*C and water's is 1 cal/g*C. The latent heat of fusion of water is 80 cal/g.

A. how much heat is required to raise the ice to 0C and completely melt the ice?

B. How much additional heat is required to heat the water (obtained by melting the ice) to 25C

C. What is the total heat that must be added to convert the 80 g of ice at -20C to water at +25C?

D. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45C? Explain.

Explanation / Answer

1)H1=150*0.5*20+150*80=13500 cal

2)H2=150*1*25=3750

3)H3=17250 cal

4)Hmelt=6400 cal

H=80*1*45=3600 cal

Htot=10000cal

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