Recent content by MixedUpCody

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

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...

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...

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