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Thermal physics hard!

  1. May 22, 2005 #1
    hi there guys im doing a thermal physics question and its hard please help me and point me in the right directions please thanks.

    question 1

    A 50g copper calorimeter contains 250g of water at 20degrees celsius how much steam must be condensed into water if the final temperature is to reach 50degrees?

    what i did i tried applying these equations
    Qcold= -Qhot
    MwCw(T-Tw) = MxCx(T-Tx)
    i let mw=0.25g

    now the tricky bit is the steam getting condensed to 50degrees i tried using the latent heat of evaporation but with no sucess and how does the copper insulator relate??? pleaes help

    Question 2
    An insulated vessel 250g if ice is at 0degrees and is added to 600g of water at 18degrees a) what is the final temperature of water

    ok first step here i calculated the energy required to change the phase of ice to water
    Q=83.3 KJ

    then i calcualted the energy 600g of water released and got Q=45.2 KJ

    how do i go from here to get final temperature of the system??? and the book says the answer is 0 how is that??? please show me thanks
  2. jcsd
  3. May 22, 2005 #2
    1. Heat gained by 250gms of water from 20deg.C to 50 deg.C is equal to the latent heat loss by 'x' gms steam at 100deg.C + sensible heat loss by 'x' gms of water from 100deg.C to 50deg.C.

    2. What will be the temperature of final mixture when you can't supply heat to melt the ice totally?
  4. May 22, 2005 #3
    1) I think that you'll have to use the relationship
    Heat gained by copper calorimeter and water=Heat lost by Steam via condensation +Heat lost by 100 degree celcius of water(after condensation) to 50 degree celcius
    2) For question 2, I think that you require the concept of thermal equilibrium. Since the energy released by the water is not enough to melt all the ice, it logically follows that some ice remains and the remaining water is also at 0 degree celsius as all of their heat ennergy is lost to melt the ice.
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