Equilibrium Temperature with three substances

[ContagiousIntellect]

Homework Statement

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 ?

Homework Equations

Q=mcΔt
Tf=(mct+mct+mct)/(mc+mc+mc)

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.

 

[Borek]

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

 

[mjc123]

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

 

[Chestermiller]

Homework Statement

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 ?

Homework Equations

Q=mcΔt
Tf=(mct+mct+mct)/(mc+mc+mc)

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.

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