# Several Questions Re. Thermodynamics

1. Dec 7, 2008

### Newtime

I'm going to post one or a few questions from a recent chemistry test on which the professor has offered extra credit if we correct the entire test. He has given us the answers to the questions but not the solutions. I have been working on the following question in particular with a few friends for almost two hours and we just can't seem to get the right answer but always very close. The answer is 35.6 deg. C.

1. The problem statement, all variables and given/known data

An ice cube at 0 degrees C weighing 9.0 g is dropped into an insulated vessel containging 72 g of water at 50 degrees C. What is the final temperature of the water after the ice has melted and a constant temperature has been reacher? The latent heat of fusion of ice is 6.01 kJ/mol, the molar heat capacity of water is 75.4 J/mol K.

2. Relevant equations

nC$$\Delta$$T = -nC$$\Delta$$T
mC$$\Delta$$T = -mC$$\Delta$$T

3. The attempt at a solution

Like I said, we have tried many different approaches but the basic reasoning behind all of them is that whatever heat the ice gains has to be lost by the water, but the once the system reaches equilibrium, the amount of heat lost will equal the amount of heat gained. I guess the main question is when to use the latent heat of fusion or if at all. Any help is greatly appreciated. Thanks.

2. Dec 7, 2008

### mgb_phys

You seem to have the correct idea.
The energy lost by the ice melting and the resulting water heating is equal to the energy lost by the hot water.
So if T is the final temperature (in centigrade)
mass_ice * latent heat ice + mass_ice_water * specific heat * T = mass_water * specific heat * (50-T)

But the numbers in, rearrange and solve for T

3. Dec 7, 2008

### Newtime

I've plugged in most conceivable numbers for the values you listed above and each time I get close but never the right answer. For mass of ice, 9g, for latent heat of ice, I used the value listed but in joules, for mass of icewater I used 81g, for specific heat I used 4.18, for mass of water I used 72g and the same specific heat as above. Am I just missing something?

4. Dec 7, 2008

### mgb_phys

How does 9g of ice produce 81g of ice water?
Perhaps I can make it a little clearer:

mass_ice * latent heat ice + mass_ice_water * specific heat * T = mass_HOT_water * specific heat * (50-T)

5. Dec 7, 2008

### Newtime

I'm not quite sure what you mean. I added the mass of ice and mass of water to get the value for "mass_ice_water." Here's the equation I used:

9.0g * (6.01E-3 J/mol) * (1 mol ice / 18.02g) + (81 g ice & water) * (4.18 J / g*K) * T = 72g * (4.18 J / g*K) * (323.15K - T)

6. Dec 7, 2008

### mgb_phys

Yes I thought that was what you were doing - it's wrong.
You don't need to consider the water mixing, the problem would work just as well if the ice was in a plastic bag when it was put in the warm water.
You are only concerned about where the energy goes - so you need to balance the water at 0deg rising to the final T and the hot water cooling to T.

9g * 6.01 kJ/9g + 9g * 75.4 J/9g * T = 72g * 75.4 J/9g * (50 - T)

7. Dec 7, 2008

### Newtime

Awesome, thank you. I have a few more questions about more or less the same thing, should I post again in this thread of make a new one for each questions? It looks like there's only 2 more.

8. Dec 7, 2008

### mgb_phys

Post in new threads, more people are likely to reply - but make an attempt at answering them!

Last edited: Dec 7, 2008