Finding the final temperature of mixed ice and steam

[kiro484]

Homework Statement

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

Homework Equations

Q=mcΔt
Q=mH

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.

 

[haruspex]

Duplicate of https://www.physicsforums.com/showthread.php?p=4167556 , https://www.physicsforums.com/showthread.php?p=4167556 , https://www.physicsforums.com/showthread.php?p=4168938, https://www.physicsforums.com/showthread.php?p=4169759
Kiro it has already been pointed out to you in several of these threads where you are going wrong. Your calculation assumes it all ends as water. The result you get proves it doesn't. So what alternative should you consider?

 

[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.

 

[ehild]

I wrote in your other thread that

The ice can not condense all steam. Steam and water coexist at the boiling point of water.

Do you understand what it means? ehild

 

[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?

 

[ehild]

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?

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

 

[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.

 

[ehild]

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.

Yes, you can do it.

Do not use so many digits in the end result. ehild

 

[kiro484]

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

 

[ehild]

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

Yes. To three digits.

ehild

 

[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.