# Finding final temperature of a mixture

## Homework Statement

calculate the final temperature when 20 grams of steam at 100°C is added to 100 grams of ice at -40°C.

Q=mcΔt and Q=ML

## 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.

tiny-tim
Homework Helper
Welcome to PF!

Hi MixedUpCody! Welcome to PF!
calculate the final temperature when 20 grams of steam at 100°C is added to 100 grams of ice at -40°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 …
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!

Hi,
Thank you for the fast response.

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

tiny-tim
Homework Helper
Qc = mcΔt + mL + mcΔt = …
= 41660 + 0.419T

Qh = mL + mcΔt = …
= -(83.72T + 36828)
when I set them equal to each other, T came out to be -932.8 °C
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?​

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

tiny-tim
Homework Helper
Hi MixedUpCody!

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

HI,

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