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Specific Heat Problem Help!

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data

    A .4lb piece of Hg at 210 F is placed in .3lb of water that has a temperature of 70.0f. The water is in .1lb brass calorimeter.

    What is the final temperature of the mixture?

    2. Relevant equations

    Q=MC(deltaT)


    3. The attempt at a solution

    The part that I'm confused from is.

    Qlost=Qgained
    What is the final and initial temperature of the silver?
    I cant figure that out from the problem, would the final temperature of the silver be 70.0degreeS?
     
  2. jcsd
  3. Mar 3, 2010 #2

    rock.freak667

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    The heat lost by the Hg = heat gained by the water + heat gained by the calorimeter

    Initially, the water and calorimeter should be at the same temperature.

    So can you formulate an expression for the given terms?
     
  4. Mar 4, 2010 #3
    I cannot formulate an expression for the given terms. I've completed algebra, adv. algebra and now im precalc and I just cant figure out the algebra to complete this practice problem.
     
  5. Mar 4, 2010 #4

    rock.freak667

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    Let's call the final temperature Tf


    using Q=McΔT

    What is the heat lost from the Hg?
     
  6. Mar 4, 2010 #5
    Well I understand that -(Qloss)=Qgained and you need to get all variables to one side and other numbers on the other side, this is what i have done but I'm stuck

    -(Qloss)=Qgained
    -(Qloss silver)=Qgained of water=Qgainedofcalirmiter
    -(mcdeltat)=mcdeltat+mcdeltat
    -(181.8181g(.214cal/gc)(tf-98.8c)=(136.3636g(1cal)(tf-21.0c)+(45.454545g)(.0917cal/gc)(tf-21.0c)

    okay this is what I have now I do not understand the algebra. ?
     
  7. Mar 4, 2010 #6

    rock.freak667

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    Can you expand the brackets and then move all the Tf to one side?
     
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