The initial temperature of 150 g of ice is -20 degree C The specific heat capaci

Question

The initial temperature of 150 g of ice is -20 degree C The specific heat capacity of ice is 0.5 cal/g middot C degree and water's is 1 cal/g middot C degree The latent heat of fusion of water is 80 cal/g. a. How much heat is required to raise the ice to 0 degree C and completely melt the ice? b. How much additional heat is required to heat the water (obtained by melting the ice) to 25 degree C? c. What is the total heat that must be added to convert the 80 g of ice at -20 degree C to water at +25 degree C? 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 45 degree C? Explain.

Explanation / Answer

Heat required is Q = Q1+Q2

Q1 is the heat required to raise upto 0 deg C

Q1 = m*Sice*dT = 150*0.5*20 = 1500 cal

Q2 = m*Lf = 150*80 = 12000 cal

then Q = 1500+12000 = 13500 cal

b) Q_additional = m*Sw*dT = 150*1*25 = 3750 cal

C) Q = Q1+Q2+Q3

Q = (m*S_ice*dT)+(m*Lf)+(m*Sw*dT)

Q = (80*0.5*20)+(80*80)+(80*1*25) = 9200 cal

D) No,we cannot find,because the amount of heat required to become water from ice is difference from the amount of heat required to reach 25 deg C from water at 0 deg C

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