The initial temperature of 150 g of ice is -20 degree C. The specific heat capac
ID: 1508373 • Letter: T
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. How much heat is required to raise the ice to 0 degree C and completely melt the ice? How much additional heat is required to heat the water (obtained by melting the ice) to 25 degree 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? 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
a) Q = m*S_i*dT + m*Lf = (150*0.5*20)+(150*80) = 13500 cal
b) Q = m*S_w*dT = 150*1*25 = 3750 cal
C) Q = 3750+13500 = 17250 cal
d) No,in the case of ice from -20 C to 0C we will use specific heat capacity of ice,where as from 0C to 25 C we use specific heat capacitity of water whose values are different from each other
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.