Latent Heat - Heat fusion of ice

[sarah-108]

The following experiment was performed to determine the latent heat of fusion of ice. A 50.0-g aluminum cup was filled with 100 g of warm water. The temperature of the cup and water stabilized at 35C. A 15-g piece of ice at 0C was placed in the water and quickly melted. The final temperature of the system was 21.4C. What value was obtained for the heat of fusion of ice?Energy (heat) = mcΔT
m= mass
c= specific heat
ΔT = change in temperature

Energy (heat) = mHf
m = mass
Hf = heat of fusion

Specific Heat of Water: 4200 J/kgC
Specific Heat of Ice: 2100 J/KgC
Specific Heat of Aluminum: 900 J/KgC

Latent Heat of Fusion for Water: 335,000 J/Kg
Latent Heat of Vaporization for Water: 2,260,000 J/KgOkay, so I'm sure the aluminum cup and the water play a role when solving for this problem, but I'm not quite sure what to do. My physics teacher hasn't taught us this yet. In fact he started our entire thermal energy unit 3 days before Christmas break, expects us to learn this on our own over the break, and we're being given a test this tuesday after 2 1/2 weeks off. Anyways, this is what I have so far:Eh (ice) = mcΔT
= (0.015kg)(2100 J/KgC)(21.4C)
= 674.1 J

Eh = mHf
Hf = Eh/m
=674.1 J/0.015Kg
= 44940 J/Kg

Thanks for anyone who could help me with this!

 

[arunma]

These calorimetry problems are actually very straightforward. Usually you're putting a sample in a calorimeter and water that have some mass and temperature, and after a few minutes the system will settle into some new equilibrium temperature. The equation you want use is:

Heat gained/lost by calorimeter & water = Heat lost/gained by sample

When matter gains or loses heat, the relation between temperature and heat is:

$$Q = mc\Delta T$$ (c is the specific heat)

When some sample of matter changes phase, the equation for the heat transfer is:

$$Q = mL$$ (L is the latent heat)

There are two tricks to remember in calorimetry problems. First of all, remember that when a sample of matter change phase, it's specific heat also changes, i.e. you can't use the same c for ice and water. So if you drop some ice with mass m in a calorimeter and it starts at some temperature $T_1 < 0^o C$, and it turns into water at temperature $T_2 > 0^o C$, then the formula for the heat transfer is NOT:

$$Q = mc(T_2-T_1)$$

Instead you have to first write the heat required to heat the ice to the freezing point, then add the heat required to change the phase into liquid water, and then add the heat required to raise the temperature from the freezing point to the new temperature. So instead the correct answer would be:

$$Q = mc_{ice}(0^o - T_1) + mL + mc_{water}(T_2 - 0^o)$$

The second trick is to make sure you get the signs right. Notice in the above equation how I placed the $0^o$ before the $T_1$ in the term for the heating of the ice, but I placed the $T_2$ before the $0^o$ in the term for the heating of the water. That was to ensure that both of these terms are positive. When we talk about the heat transferred by the calorimeter & water and the heat transferred by the sample, we're talking about positive values. So when you substitute the initial and final temperature values into the $\Delta T$'s, you want to put them in the proper order so that each $\Delta T$ will be positive. I've had lots of students who got these problems mostly right but screwed up on this point, so it's quite a common error. Sometimes the problem won't tell you the initial or final temperature, so you need to pay attention to whether the sample that's being dropped in is hotter or colder than the water bath of the calorimeter, and order the temperatures accordingly. Alternatively, you can subtract the final temperature from the initial temperature when writing $\Delta T$ no matter what, but place a positive or negative sign in front of the whole term accordingly, but I personally find it easier to just put it in the $\Delta T$.

So for this problem, write the heat transferred from the calorimeter and water bath to the ice. Also write the heat transferred from the ice to the rest of the system (remember the phase change!), and then set the two expressions equal to each other. You should now be able to calculate the latent heat of fusion of ice, and you can even compare it to the textbook value (which you could Google) to make sure you got a reasonable answer.

 

[kerimek]

Hi there! It seems like you're on the right track with your calculations. Let's go through the steps to solve for the heat of fusion of ice.

First, we need to calculate the energy that was transferred from the warm water to the ice in order to melt it. This is given by the equation: Eh (ice) = mcΔT, where m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature.

You correctly calculated the energy transferred to the ice as 674.1 J. Now, we need to use this value to solve for the heat of fusion of ice.

The heat of fusion is defined as the amount of energy required to change 1 kg of a substance from solid to liquid at its melting point. In this case, we have 15 g of ice (0.015 kg) that has been melted, so we need to divide the energy transferred by the mass of the ice to get the heat of fusion.

Hf = Eh/m = 674.1 J/0.015 kg = 44940 J/kg

This means that it takes 44940 J of energy to melt 1 kg of ice at its melting point. This value is close to the accepted value of 335,000 J/kg, so your experiment was successful in determining the heat of fusion of ice.

Hope this helps and good luck on your test!