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Homework Help: Thermodyanmics, specific heat capacity

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    a 4kg iron hammer is initally 700 degree celcius is droppped into a bucket containing 20kg of water at 25degree celcius. what is the final temperture



    2. Relevant equations

    q=mc(delta)T

    3. The attempt at a solution

    m1=4kg
    m2=20kg
    T1=700=973K
    T2=25=298
    i know the system is in equilibrium but i dont know how to solve it correctly.. so heres my shot at it

    Q=mc(delta)T
    =(4)(449J/kgK)(Tf-973)=(20)(4190J/kgK)(Tf-298)

    i think i'm doing this question wrong because .. when i solve for Tf i get a negative number of -288.66K

    help?
     
  2. jcsd
  3. Nov 27, 2008 #2
    nvm forget this question i just figured out how to do it... i forgot to put the negative sign for the iron since it loses heat.
     
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