# Finding final temperature of a mixture

1. May 18, 2012

### MixedUpCody

1. The problem statement, all variables and given/known data
calculate the final temperature when 20 grams of steam at 100°C is added to 100 grams of ice at -40°C.

2. Relevant equations
Q=mcΔt and Q=ML

3. The attempt at a solution
I tried solving this problem using -ΔQheat = ΔQcold, but the final temperature that I found was past 100 °C which is impossible because the steam's temperature is only at 100°C. I believe I need to find how much of the ice actually turns into water and how much of the steam actually condenses into water but I don't know how to go about doing that. Thank you for the help.

2. May 18, 2012

### tiny-tim

Welcome to PF!

Hi MixedUpCody! Welcome to PF!
No, you can assume that all the steam and ice ends up at the same temperature and in the same state.

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

3. May 18, 2012

### MixedUpCody

Hi,
Thank you for the fast response.
This is what I had:

Qc = mcΔt + mL + mcΔt = (0.1)(2090)(0-(-40)) + (0.1)(3.33 x 10^5) + (0.1)(4.19)(T-0)
= 41660 + 0.419T

Qh = mL + mcΔt = -[(0.02)(22.6 x 10^5) + ( 0.02)(4186)(T-100)]
= -(83.72T + 36828)
when I set them equal to each other, T came out to be -932.8 °C

4. May 18, 2012

### tiny-tim

no, that minus shouldn't be there

the first one is energy gained by the ice

the second is energy lost by the steam …

they should both be positive, shouldn't they?​

5. May 18, 2012

### MixedUpCody

that's true. I don't know why I'm hung up on the negative from the -ΔQh = ΔQc equation. If I don't put the negative there the answer is 58°C, but the answer my teacher gave me in class was 23.4°C. The teacher didn't really show us how to get that answer,. Would you mind helping me with this? Thank you

6. May 19, 2012

### tiny-tim

Hi MixedUpCody!

(just got up :zzz:)
why are you using 4.19 and 4186 for the same specific heat?

7. May 20, 2012

### MixedUpCody

HI,

I see...lol..I'll try again with the same number =)..thank you