If she is served 220 mL of coffee at 80 C in a well-insulated container, how muc

Question

If she is served 220 mL of coffee at 80 C in a well-insulated container, how much ice at 0 C should she add to obtain a final temperature of 39 C

mass of coffee = 220 grams = 0.22 kg specific heat of coffee(water) = 4186 J/(kg K)

let m is the mass of ice that should be added. Lf = 3.33*10^5 J/kg

Apply, heat lost by coffee = heat gained by ice

0.22*4186*(80 - 39) = m*Lf + m*C_water*(39 - 0) 37757.72 = m*( 3.33*10^5 + 4186*39) m = 37757.72/( 3.33*10^5 + 4186*39) = 0.076 kg or 76 grams

I understand how to solve the problem, but why is it that I have to use the specific heat of water and not ice when calculating the heat gained by the ice?

Explanation / Answer

mass of coffee = 220 grams = 0.22 kg

specific heat of coffee(water) = 4186 J / kg K

let m is the mass of ice that should be added.

Lf = 3.33 X 105 J / kg

Apply, heat lost by coffee = heat gained by ice

0.22 X 4186 X (80 - 39) = m X Lf + m X C water X (39 - 0)

37757.72 = m X ( 3.33 X 105 + 4186 X 39)

m = 37757.72 / ( 3.33 X 105 + 4186 X 39)

m = 0.076 kg or 76 grams

when calculating the heat gained by the ice

actual thing is happening that once ice starting osorbing heat means

at that perticular instant ice will be converted into water.

either it is cool water or what ever temperature of water may be.

so to calculate the heat gained by ice means compulsary have to use

specific heat of water

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