1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equilibrium Temperature

  1. Nov 24, 2004 #1
    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 [tex]Q = mc \Delta T[/tex]
    -(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. jcsd
  3. Nov 24, 2004 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  4. Nov 24, 2004 #3
    oh yeah and i also didn't need the negative sign. i got it thanks
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Equilibrium Temperature