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).
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.
Heat that water lost is: Qlost = mwater.cwater.(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 : Qgained=mice.Lf = mice. 333
Heat that 0°C water gained to rise its temperature to 1°C is: mice.cwater.(1°C - 0 °C) = mice.4.19
So the total heat that ice gained to rise from 0°C ice to 1°C water is: Qgained=333mice+4.19mice=337.19mice
Because Qgained=Qlost, so 337.19mice=8.38 → mice=8.38/337.19=0.024852...≈0.025 kg.
Q = m.Lf=m.c.Δt,
in which Lf 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).
mice = Q/Lf = mwater c Δt /Lf = 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.