Science Buddies: "Ask an Expert"

if a 100g piece of copper at 80 degrees celsius is placed into 200g of water at 20 degrees celsius, what will the final equillibrium temperature be? (specific heat capacity of copper = 39J/Kg degrees celsius, and specific heat capacity of water is 4182 J/Kg degrees celsius.)

So far, I have worked out
Q= m*c*△θ

Say final temperature of the copper is 'x', and final temperature of the water is 'y'.

Q(copper) = 200g * 4182 * (x-80)
Q(water) = 200g * 4182 * (y-20)

100g * 39 J/Kg * (x-80) = 200g * 4182 J/Kg * (y-20)

This, so far, is where my teacher says I am correct. I have no idea what I'm doing wrong.

3900 * x - 31200 = 836400 * x * -16728000

(3900 - 836400)x = -16416000 + 312000

-832500 * x = -164160000

x = -16416000/-832500
x = 19.71891892
x = 19.7 (3 sf)

Hi - this forum is for students doing science projects, we are unable to help with homework questions.

My apologies, thank you anyways.