Finding the final temperature of mixed ice and steam

1. Nov 25, 2012

kiro484

1. The problem statement, all variables and given/known data
Suppose that 20.0g of steam at 110°C is mixed with 25.0g of ice at -40°C What will the final temperature be?

Ice - Specific heat capacity 2.10 J/g°C
Water - Specific heat capacity 4.19 J/g°C
Water - Latent heat of fusion 334 J/g
Water - Latent heat of vaporization 2268 J/g
Steam - Specific heat capacity 2.08 J/g°C

2. Relevant equations
Q=mcΔt
Q=mH

3. The attempt at a solution
Heat loss (steam) = Heat gain (ice)
mcΔt+mH+mcΔt = mcΔt+mH+mcΔt
(20g)(2.08J/g°C)(10°C)+(20g)(2268J/g)+(20g)(4.19J/g°C)(100-t) = (25g)(2.1T/g°C)(40°C)+(25g)(334J/g)+(25g)(4.19J/g°C)(t-0)
416J+45360J+83.8J/°C(100-t) = 2100J+8350J+104.75J/°C(t-0)
Combining like terms
45776J+83.8J/°C(100-t) = 10450J + 104.75J/°C(t-0)
35326J+83.8J/°C(100-t) = 104.75J/°C(t-0)
35326J+8380J/°C-83.8J/°Ct = 104.75J/°Ct
35326J+8380J/°C = 188.55J/°Ct
187.356139(1/°C)+44.444444444444444444444444444444 = t
231.8005834°C = t
232°C = t

Any help would be greatly appreciated, thanks in advance.

2. Nov 26, 2012

haruspex

Last edited by a moderator: May 6, 2017
3. Nov 26, 2012

kiro484

It ends in 2 different states. But I don't know how to show any of the work needed to show this. Am I supposed to say that only (for example, not the actual values) that only 15g of steam condenses? What about the other 5g? Does it stay at 110 degrees? It can't do that because then you have 2 different temperatures. This is what I don't understand, and I've asked many times hoping more people will be able to help me.

4. Nov 26, 2012

ehild

Do you understand what it means????

ehild

5. Nov 26, 2012

kiro484

Ok so that means they both have to be at 100 degrees. All the steam cooling to that temperature releases 416J of heat. The ice needs 21344J to reach 100 degrees. Where would the rest of this energy come from? Some of the steam condensing? How much of the steam would be needed to provide that heat?

6. Nov 26, 2012

ehild

It is some steam condensing. So use t=100 in your equation, but when multiplying the latent heat of evaporation with the mass x instead of 20 g.

ehild

7. Nov 26, 2012

kiro484

Could I do this? This is using the values of ice needing 21344J to reach 100 degrees and steam releasing 416J when becoming 100 degrees.
21344-416=20928J needed to get all the ice to 100 degrees.
Q=mH
m=Q/H
m=20928J/2268J/g
m=9.22751323g
So that is the mass of steam needed to condense to release the rest of the energy needed to bring the ice up to 100 degrees.

Last edited: Nov 26, 2012
8. Nov 26, 2012

ehild

Yes, you can do it.

Do not use so many digits in the end result.

ehild

9. Nov 26, 2012

kiro484

Yeah you have to round off to the amount of significant digits of the original values given in the question, correct?

10. Nov 26, 2012

ehild

Yes. To three digits.

ehild

11. Nov 26, 2012

kiro484

Ok thank you so so much for all the help. Sorry for creating a billion threads trying to find more help, yours is all I needed.