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## Homework Statement

You take a 0.5 L bottle of water (5 °C) from a refrigerator, and pour it into a glass. Find the mass of ice at 0 °C needed to cool your drink down to 1 °C ? the heat of fusion of water is 333 kJ/kg, and the specific heat of water is 4.19 kJ?(kg.°C).

## Homework Equations

## The Attempt at a Solution

This is a very easy problem. But somehow my solution is different from the solution in my textbook. Therefore I would like you to point out what is my misunderstanding here.

My solution:

Heat that water lost is: Q

_{lost}= m

_{water}.c

_{water}.(5°C - 1°C) = 0.5x4.19x(5-1) = 8.38 kJ.

Heat that ice gained to change from 0 °C ice to 0 °C water is : Q

_{gained}=m

_{ice}.L

_{f}= m

_{ice}. 333

Heat that 0°C water gained to rise its temperature to 1°C is: m

_{ice}.c

_{water}.(1°C - 0 °C) = m

_{ice}.4.19

So the total heat that ice gained to rise from 0°C ice to 1°C water is: Q

_{gained}=333m

_{ice}+4.19m

_{ice}=337.19m

_{ice}

Because Q

_{gained}=Q

_{lost}, so 337.19m

_{ice}=8.38 → m

_{ice}=8.38/337.19=0.024852...≈0.025 kg.

Textbook solution:

Q = m.L

_{f}=m.c.Δt,

in which L

_{f}is heat of fusion of the transition from ice to water is 333kJ/kg, and c is the specific heat of water, 4.19 kJ/(kg.°C).

m

_{ice}= Q/L

_{f}= m

_{water}c Δt /L

_{f}= 8.38/333 = 0.02512651...≈0.025kg.

We see that textbook solution does not include

**the heat that 0°C water gained to rise its temperature to 1°C**. Why is that? I believe that whatever temperature is, as long as it's above zero, things need to gain some certain amount of heat to rise its temperature from zero °C to that temperature.