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I'm over-complicating thermal equilibrium HELP PLEASE.

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2kg block of ice initially at T = −20 ◦C is put in contact with a very large amount of liquid water at T = 0 ◦C (by “very large amount” we imply that by the time ice and water reach thermal equilibrium, there will still be some liquid water left). Ice and water can exchange heat just with each other but not with the environment. The specific heat of water is 4186 J and of ice is 2050J.



    1)final equilibrium
    2)amount of heat absorbed by the ice in the process
    3)final mass of the ice at equilibrium

    2. Relevant equations
    deltaEbond + deltaEthermal = Q + W
    deltaEthermal = m•Csp•deltaT
    deltaEbond=|deltamass•deltaH|
    deltaH(meltin, ice)=6.01kJ/mol


    3. The attempt at a solution
    My first guess is that the equilibrium temperature is 0ºC, because we're to assume that water is still present.

    assuming thats correct could I then use deltaEthermal = m•Csp•deltaT = 2kg • 2050J/kg•K • (0ºC-(-20ºC)) = 82kJ
    and then, assuming that all the ice melts, calculate deltaEbond = |(mf - mi)•deltaH(melting, ice)| = |(0kg - 2kg)•6.01| = 12.02kJ

    then, since Q=deltaEthermal + delta Ebond, Q = 82kJ + 12.02kJ = 96.02kJ

    But I had to assume there was enough E to melt all the ice to calculate this... and one of the questions asks to calculate the amount of ice left!

    I feel fairly stupid for not being able to do this, I think I'm waaaaaaaay over-complicating it. Thanks internet!!!!
     
    Last edited: Mar 14, 2010
  2. jcsd
  3. Mar 14, 2010 #2

    ehild

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    It is enough water, so there will be some liquid water at the end of the process -at 0°C! From where does the ice get heat to warm up to 0°C? What happens to 0°C water if you take away some heat of it? Can the ice melt? So what happens to the water?

    Take care on the data: deltaH(melting, ice)=334 kJ/kg.

    ehild
     
  4. Mar 14, 2010 #3
    Sorry, I had deltaH in moles.

    So I see what you're saying; all the heat that goes to warming the ice to 0ºC comes from the water, which in turn freezes portions of it. So step one, calculate the energy it takes to raise 2kg of ice to 0ºC, then negate that answer and set it equal to the bond energy of freezing a certain amount of water and solve for the difference. This is how much ice is added to the 2Kg block of ice.

    deltaE(thermal,ice) = (2kg)•(2050J/kg•K • (0ºC-(-20ºC)) = 82kJ
    deltaE(bond, water) = -82kJ = (mi - mf)•334kg => -.245kg = mf - mi
    m(ice, final) = 2kg + .245kg = 2.245kg of ice at equilibrium

    I hope that's right.
     
  5. Mar 14, 2010 #4

    ehild

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    Correct!

    ehild
     
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