# Ah temp help

1. May 2, 2005

### oldschool

got a proplem that i jsut can't seem to get
200g of water is placed in a 80g aluminum cup and left undesturbed in a freezer until it reaches -5c the cup is then taken out and heated to 96c
a. how much heat is needed to increase the temp (from the -5c to 96c)
b. replace the alum cup with an iron cup of 120g and replace the water with ethyl alcohol how much alcohol is neded to add the same amount of heat as in part A and to raise the temp from -5c to 96c

thanks for any help
john

2. May 2, 2005

### Chi Meson

a) use the "specific heat" formula: Q = cm delta-T. But use it twice. FInd the total heat (Q_cup + Q_water) for the same change in T.

b) once you have the total Q, set up the same equation as before (except now for the specific heats of iron and alcohol). Use the total Q from part a and solve for the mass of alcohol.

3. May 2, 2005

### oldschool

so for a it would be Q=(200g)(.5)(-5-0) (to change from ice to a liquid) + (200g)(79.7)(latent heat)+(200g)(1)(0-96)

do i jsut add on the aluminum q then to that ?

Qalum=(80g)(.22)(-5-96)

looking at it i belive i do jsut add that on but i would like to make sure im rite lol

4. May 2, 2005

### minger

Well, you don't simply have to increase the temperature. At 0°C, you have to melt the ice. That will require some amount of kJ/kg. Only have that heat is thrown into the ice, can you start to raise the temperature of the water some more.

So in addition to Qrequired to heat the water and Qrequired to heat the can, you also have Qrequired to melt the ice.

5. May 2, 2005

### oldschool

then for b it would be (assuming the top is correct and i did cacls right it should be 37417.6 cals)

37417.6=(m)(.58)(-5-78)+(m)(25)+(m)(.58)(78-96)

that look correct?

6. May 2, 2005

### OlderDan

What you meant to say was

so for a it would be Q=(200g)(.5cal/g.oC)(0-(-5))oC + (200g)(79.7cal/g) (latent heat; to change from ice to a liquid))+(200g)(1cal/g.oC)(96-0)oC

and

Qalum=(80g)(.22cal/g.oC)(96-(-5))oC

Here i assume you are talking about the alcohol, and I will assume you have the constants correct, except for the missing units (we all do that, but we shouldn't) and the temperature values being reversed again (this is what I wanted to call attention to more than the units). Of course you still have to find the heat for the cast iron cup. Plus, there might be a bit of a "trick" to this question. If the cast iron cup is open, how much heat is it going to take to raise the temperature of the cup above the vaporization temperature of the alcohol? Where is the alcohol vapor?

Last edited: May 2, 2005
7. May 3, 2005

### oldschool

thanks for catching me on the iron cup lol i forgot all about that i saw the proplem of revered temps when i did the equation so i fixed that the vaper point of alcohol is like 118 i think so thankfully i didn't ahve to worry about that and the cup wasn't supposed to loose any heat

thanks for the help though

8. May 3, 2005

### Chi Meson

NB:
yes I did forget about the latent heat of ice melting. But someone already caught it. Ah, well.