# Equilibrium Temperature

1. Nov 24, 2004

### mikep

A 199 g lead ball at a temperature of 80.9°C is placed in a light calorimeter containing 178 g of water at 24.5°C. Find the equilibrium temperature of the system.

i used the equation $$Q = mc \Delta T$$
-(199g)(4.186J/g°C)(T - 80.9°C) + (178g)(0.13J/g°C)(T - 24.5°C) = 0
T = T final = 82.5°C
can someone please tell mw what i did wrong?
can someone also confirm if the specific heat for lead is 0.13J/g°C?

2. Nov 24, 2004

### Tom Mattson

Staff Emeritus
You put the specific heat of water in the term for the lead ball, and vice versa.

Not off the top of my head, but if you type "specific heat lead" into Google, I'm sure you can find it.

3. Nov 24, 2004

### mikep

oh yeah and i also didn't need the negative sign. i got it thanks