# Equilibrium Temperature with three substances

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1. Jan 26, 2017

### ContagiousIntellect

1. The problem statement, all variables and given/known data
If a block of 250 grams of lead (specific heat (c) =130) at 315 degrees celsius is placed in a 200 gram aluminum (c=900) calorimeter cup containing 900 grams of water (c=4.1), and the calorimeter and the water are both initially at 15 degrees celsius, what is the equilibrium temperature ?

2. Relevant equations
Q=mcΔt
Tf=(mct+mct+mct)/(mc+mc+mc)
3. The attempt at a solution
Tf=((250x900x315)+(200x900x15)+(900x4.1x15))/((250x130)+(200x900)+(900x4.1)
Tf=(70875000+2700000+55350)/(32500+180000+3690)
Tf=40.1
I'm not sure if I'm on the right track.

Last edited: Jan 27, 2017
2. Jan 27, 2017

### Staff: Mentor

Try to approach it in a systematic way - for a single substance Q=mcΔt, for the whole system

$$\sum_i m_ic_i\Delta T_i = 0$$

where $\Delta T_i = T_{final} -T_{initial(i)}$ (Tfinal is common for all substances present).

3. Jan 27, 2017

### mjc123

What are the units of specific heat, if lead is 130, aluminium is 900 and water is 4.1???

4. Jan 28, 2017

### Staff: Mentor

The math is not consistent with your second Relevant equation. Try again.