# Specific latent heat of fusion of ice question

• maxim07

#### maxim07

Homework Statement
In an experiment to measure the specific latent heat of fusion of ice, 20g of crushed ice is added to a beaker of water of mass 160g (not including mass of beaker) at temperature 30°C. The final steady temperature of the water and ice mixture is found to be 20°C.
Take c to be 4.2 Jg^-1°C^-1

a) calculate the specific latent heat of fusion of ice in J/g

b) if 3g of the ice has already melted before it is added to the beaker, show how this changes the value for the latent heat of fusion.

I can do (a) but I know that the value for the latent heat of fusion in (b) should be smaller as less energy is needed to melt the ice, as some has turned into water. But the result I get for b is larger than for a. Is it correct?
Relevant Equations
Energy transferred from water = energy to melt ice + energy to raise temperature of ice that is now water

Ew = EL + Ei
a)
Ew = EL + Ei

mwc(T2-T1) = miL + mic(T2-0)

160 x 4.2 x (30 - 20) = 20L + (20 x 4.2 x 20)

L = (6720 - 1680)/20

= 252 J/g

b)

accounting for 3 gram of melted ice

160 x 4.2 x (30 - 20) = (20 - 3) + (20 x 4.2 x 20)

L = (6720 - 1680)/17

= 296 J/g

I think that your solution is correct.
Case I:- x is the amount of energy required to melt y gram of ice and then raising it's temprature to z.
Case II:- x is the amount of energy required to melt k(<y) gram of ice +raising temprature of y gram of water to z .
Now We can see that same amount of energy is required to melt ice in case 1 and in case 2 and we know that amount of ice in case 2 is less than that of case 1 so now it seems obvious that latent heat of fusion of ice in case 2 should be more than that of case 1 as it's amount is less.

but I know that the value for the latent heat of fusion in (b) should be smaller as less energy is needed to melt the ice, as some has turned into water.
The energy you know is what is supplied by the warm water. If there's less ice to melt with the same heat then the latent heat must be higher.

Btw, I note you are not told the ice starts at 0C, but you have to assume that.

• Frigus
Yes that seems correct. But my textbook states that if the ice is not completely dry the value for the specific latent heat will be too small. As the specific latent heat of fusion of ice is 336 J/g, so I suppose both values I have calculated are too small as the question has accounted for experimental errors. Maybe I just have to state that the value in b is too small compared to the true value because less energy is required to melt the ice, although it is higher than that in a because in calculation b I have kept the temperature change the same. In reality would the temperature change be smaller as less energy would be required?

Yes that seems correct. But my textbook states that if the ice is not completely dry the value for the specific latent heat will be too small. As the specific latent heat of fusion of ice is 336 J/g, so I suppose both values I have calculated are too small as the question has accounted for experimental errors. Maybe I just have to state that the value in b is too small compared to the true value because less energy is required to melt the ice, although it is higher than that in a because in calculation b I have kept the temperature change the same. In reality would the temperature change be smaller as less energy would be required?
It depends whether you know 3g has already melted.
In my original post, I assumed you knew, but the rest of the data remained the same. Presumably you did the same in your calculation.
But if the experimenter is unaware that 3g has already melted then a different calculation will be performed. Since the final temperature will end up higher than it should the latent heat will be underestimated.

• Frigus
for this question the experimenter is unaware that 3g has melted, that is why I expected the latent heat to be smaller. I don’t know what calculation needs to be done to account for the 3g that has melted?

The actual heat balance will be
160 x 4.2 x (30-20) = 17Ltrue + 20 x 4.2 x 20
Ltrue = (6720 - 1680)/17
If you are unaware that 3g has melted, you will calculate it as
160 x 4.2 x (30-20) = 20Lapp + 20 x 4.2 x 20
Lapp = (6720 - 1680)/20
Lapp = 17Ltrue/20
Note that on this assumption the L value you calculate will be the same in cases a and b, but in case a it will be the true value, and in case b it will be less than the true value. (These can't both be correct, of course; I suppose the idea is that you calculate a value as in case a, and then consider how your value would be related to the true value if it turned out that some of your ice was already melted.)

for this question the experimenter is unaware that 3g has melted, that is why I expected the latent heat to be smaller. I don’t know what calculation needs to be done to account for the 3g that has melted?
Can you calculate what the final temperature would be if 3g has melted? For this, you would use the true value for ##L##. Then, what would the experimenter calculate for ##L## assuming that the experimenter is unaware that 3g has melted.

Taking true value as 334 J/g

calculating new final temperature

160 x 4.2 x (30 - T2) = 17 x 334 + (20 x 4.2 x T2)

20160 -6720T2 = 57112 + 84T2

756T2 = 14448

T2 = 19.1 °C

but if I put 19.1 °C into the equation with 20g as the mass, if experimenter does not know has melted, then I get L = 285.6 which is still higher than what I got in a.

As the question wants me to show how it effects my answer in a, I think the value for b needs to be lower.

There's a complication due to the fact that the experimental value of L for part (a) is not very close to the true value of L. So, there is significant "experimental error" in the experiment. You can check that if the final temperature for part (a) had been given to be 17.8 oC, the experimental value for L would have been close to the true value. Then your calculation in the previous post would show that if 3g of ice is premelted, the final temperature would rise to 19.1 oC and the calculated value of L would be lowered from the true value to 286 J/g.

So, normally, if some of the ice has premelted, the calculated value for L will come out too low, as you expected.