# Solve Latent Heat of Fusion Homework Problem

• stephenranger
L bottle of water (5 °C) from a refrigerator, and pour it into a glass.You need to cool your drink down to 1 °C by adding ice cubes of 0 °C. The mass of each ice cube is 10g.The heat of fusion of water is 333 kJ/kg, and the specific heat of water is 4.19 kJ?(kg.°C).f

## 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).

## 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: 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.

Textbook solution:

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.

## 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).

## 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: 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.

Textbook solution:

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.

I don't know why the textbook does not include that, but you're right. The heat needed for the water to go from 0 °C to 1 °C should be included. The only reason the two answers are the same is because the heat required to increase the temperature of the water is negligible as compared to the heat required to melt the ice.

• Chestermiller
If this is their mistake, then it is not a joke because the above problem is taken from a university entrance exam. I'm convinced that I misunderstood something about the problem, rather than a mistake of the exam.

The textbook's answer is wrong. It shouldn't be 8.38/333 it should be 8.38/337.19 like you said.

People make mistakes, even exam creators.

Maybe they were thinking that the glass is dipped into an ice/water bath.

Maybe they were thinking that the glass is dipped into an ice/water bath.

Yes sir actually I'm not sure about that. Let me write the full problem here because actually I adjusted it a little bit as the main problem to solve still comes down to calculating the mass of ice.

You take a 0.5 L bottle of water (5 °C) from a refrigerator, and pour it into a glass. How many ice cubes, 0 °C, do you need to cool your drink down to 1 °C, if the mass of each ice cube is 10g ? the heat of fusion of water is 333 kJ/kg, and the specific heat of water is 4.19 kJ?(kg.°C).

Yes sir actually I'm not sure about that. Let me write the full problem here because actually I adjusted it a little bit as the main problem to solve still comes down to calculating the mass of ice.

You take a 0.5 L bottle of water (5 °C) from a refrigerator, and pour it into a glass. How many ice cubes, 0 °C, do you need to cool your drink down to 1 °C, if the mass of each ice cube is 10g ? the heat of fusion of water is 333 kJ/kg, and the specific heat of water is 4.19 kJ?(kg.°C).
Well that pretty well locks it in: you were correct and they were not correct.

Well that pretty well locks it in: you were correct and they were not correct.

Excellenté sir. Thanks