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Removing heat energy from a block of ice

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    How much energy must be removed from a 4.50 cm x 4.30cm x 6.20cm block of ice to cool it from 0C to -31.0C?

    The density of ice is 920 kg/m^3. You might need to refer to the textbook for other physical constants of ice.


    2. Relevant equations

    Q = Mc(T_f-T_i)

    heat energy = mass*heat capacity * change in temp

    M = Density*volume

    3. The attempt at a solution

    Q = Mc(T_f-T_i) = DVc(T_f-T_i)

    Q = (920)(.045*.042*.062)(2180)(-31-0)= -7285.5 joules
     
    Last edited: Apr 28, 2010
  2. jcsd
  3. Apr 28, 2010 #2

    kuruman

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    Is there a question you wish to ask?
     
  4. Apr 28, 2010 #3
    Are you saying that this is right?
     
  5. Apr 28, 2010 #4

    kuruman

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    I am asking you if you wish to ask a question because your original posting does not include one. If you do, please ask and I (we) will do my (our) best to answer it.
     
  6. Apr 28, 2010 #5
    Don't you need to know how much heat you need to pull out of it? Doesn't the method I have listed do that? It finds the heat energy removed doesn't it?
     
  7. Apr 28, 2010 #6

    kuruman

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    Yes, yes and yes. However, if this is an answer that you need to put in a computerized homework system, you need to understand the sign conventions.

    If the question asks to find "the heat that goes in" and the number is negative, that means that heat actually goes out. In this case, the question is to find the heat that goes out, therefore a positive number is in order.
     
  8. Apr 28, 2010 #7
    So its just 7285.5 joules then?
     
  9. Apr 28, 2010 #8

    kuruman

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    That would appear to be the case, but I am not the program that makes the decision whether you are right or wrong.
     
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