# Homework Help: Grade 11 physic problem; calculating heat transfers?

1. Nov 30, 2011

### raininggently

1. The problem statement, all variables and given/known data
What will be the equilibrium temperature when a 245-g block of copper at 285° c is placed in a 145 g aluminium calorimeter cup containing 825 g of water at 12°c?

2. Relevant equations

q=mcΔt

qlost+qgained=0
-qlost=qgained
-m1c1t1=m2c2t2

3. The attempt at a solution
So this is what I did so far:

Know :

Copper Aluminium
- 245 g= 0.245 kg - 145 g= 0.145 kg
- specific heat capacity of copper = 0.385 j kg - ti=12°c
- ti ( Inital temp)= 285°c -teq=?
-teq=?

Water
-825 g=0.825 kg
-ti=12°c
-teq=? ( T equilibrium )

So I know qlost+ qgained= 0

then - qlost=qgained.

In this case the qlost will be the copper.

So this is what I used: ( Cu- copper; al-aluminium; & w- water)

Mcu(Ccu)(Ticu-Teq)=Mal(Cal)(Teq-Ti) + Mw(Cw)(Teq-ti)

(0.245 kg)(0.385)(285°c-Teq)=(0.145 kg)(9.2x10^2)(Teq-12°c)+(0.825kg)(4.18x10^3)(teq-12°c)

Now when I come to this, I have no clue what to do :

(0.094325)(285°c-Teq)=(133.4)(Teq-12°c)+(3448.5)(teq-12°c)

Please let me know what I`m doing wrong, and how i can solve it completely.

2. Nov 30, 2011

### Staff: Mentor

It looks like you're using J/gram/°C for the specific heat of copper, but J/kg/°C for the others. Make sure that you use consistent units for all the values.

As for what to do with the equation once it's written, I'm afraid it needs to be multiplied out and the the Teq terms collected. You might find it easier to do it symbolically, then plug in numbers afterwords.

3. Nov 30, 2011

### Barakn

You didn't do anything wrong, you just stopped right before the finish line. You now need to use the Distributive Law of algebra on both sides, collect like terms, and divide. I'll start the left side for you. (0.094325)(285°C-Teq) = 26.88°C - .094325 x Teq = ....