Thermal equilibrium, find final temp. or system?

[gibson101]

I solved the problem and have the correct answer just not sure about something.

Question:What will be the equilibrium temperature when a 274 g block of copper at 317°C is placed in a 137 g aluminum calorimeter cup containing 829 g of water at 13.0°C?

So heat lost by copper = heat gained by aluminum and water
So mcΔT= mcΔT + mcΔT

(274g)(.093 cal/g*C)(317-Tfinal)=(137g)(.22 cal/g*C)(Tfinal-12.6)+(829g)(1 cal/g*C)(Tfinal-12.6)

Solving this equation gives me 21.75 degrees C.
But why am i subtracting the final temperature from the initial temperature for copper? Whereas for the aluminum and water I am subtracting the initial from the final temp? I though ΔT was final temp - initial temp?

 

[rock.freak667]

Solving this equation gives me 21.75 degrees C.
But why am i subtracting the final temperature from the initial temperature for copper? Whereas for the aluminum and water I am subtracting the initial from the final temp? I though ΔT was final temp - initial temp?

Yes it is. But on the left side is the heat lost while on the right side the heat is gained.

If you just checked heat before = heat after then the left side would be

-(274g)(.093 cal/g*C)(T_final-317)

Note the minus sign to indicate a loss in heat.

 

[gneill]

You're balancing an equation. On one side you have heat losses and on the other heat gains. In order to equate their magnitudes you need to change the sign of one of the ΔQ's. This is accomplished by choosing the order of initial versus final temperatures accordingly. Usually the order is chosen to produce positive numbers for change in heat for comparison. So we have Q amount of heat given up on the left hand side equals Q amount of heat taken up on the other.