1. Not finding help here? Sign up for a free 30min 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!

Thermodynamics: Q=mc(deltaT)

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    In a physics lab experiment a student immersed 191 one-cent coins (each having a mass of 3.00 g) in boiling water, at a temperature of 100 degrees C. After they reached thermal equilibrium, she fished them out and dropped them into an amount of water of mass 0.275 kg at a temperature of 17.0 degrees C in an insulated container of negligible mass.

    What was the final temperature of the coins? (One-cent coins are made of a metal alloy - mostly zinc - with a specific heat capacity of 390 J/(kg * K).)

    Use 4190 J/(kg* K) for the heat capacity of the water.

    2. Relevant equations

    Q=mc(deltaT)

    Q1 + Q2=0



    3. The attempt at a solution

    I'm not sure where you can get a workable system of equations from this problem. I tried finding the Q of the coins, which I found to be Q=(191*.003kg)(390)(T-100) where T=the final temperature of the coins. And the Q of Water: Q=(.275)(4190)(T-17).

    You also know that the Q of water and the Q of the coins must be equal to 0 because they are in an insulated container.

    Because the water is heated and the coins are cooled, you know that the Q for water is positive and that Qwater=Qcoins, However, they have different final temperatures and I'm not sure how to find one in terms of the other so I can solve the equation I've come up with. Please help!
     
  2. jcsd
  3. Nov 16, 2008 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I don't know if you did it on purpose or by accident, but you used the same letter T in both equations. Is it really the same? If not, use a different letter. But if you can argue they are, you can set Qwater = Qcoins which gives you: 191*0.003*390*(T - 100) = 0.275*4190*(T - 100) and solve for T (one equation, one unknown).
     
  4. Nov 16, 2008 #3
    You can't assume the final temperatures are the same, sorry. So Qcoins=Qwater but
    (191*.003)(390)(R-100)=.275(4190)(T-17)
     
  5. Nov 17, 2008 #4
    I guess you can assume the final temperatures are equal. I think it was a problem with my signs in that I overcompensated for the fact that the coins lose heat, so they have a negative value for Q. The final temperature will obviously be less than 100 so I suppose that accounts for the sign change.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Thermodynamics: Q=mc(deltaT)
  1. Basic Thermodynamics Q (Replies: 2)

Loading...