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Heat transfer equation (temperature difference)

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    For the equation mc[tex]\Delta[/tex]T, i suppose [tex]\Delta[/tex]T should be Tf - Ti.

    However, at what condition we should let [tex]\Delta[/tex]T be Ti - Tf?

    I did come across a question which requires [tex]\Delta[/tex]T be Ti - Tf for a heat lost by an object.

    eg,
    heat lost by copper = heat absorbed by beaker & water
    mc(Ti - Tf) = mbeakercbeaker(Tf - Ti) + mwatercwater(Tf - Ti)

    I think perhaps we need to let mc[tex]\Delta[/tex]T as a positive value since it just a magnitude? however, the value is not the same(pretty close) if i let it be Ti - Tf.

    Need help, thanks =)
     
  2. jcsd
  3. Nov 17, 2009 #2
    anyone knows?
     
  4. Nov 17, 2009 #3

    Mapes

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    [itex]mc\Delta T=mc(T_f-T_i)[/itex] is always the energy gained. The energy lost by the copper was then [itex]-mc\Delta T[/itex], which is equivalent to the other expression.
     
  5. Nov 17, 2009 #4
    however, for the example above, why the initial temp and final temp is reversed?
     
  6. Nov 17, 2009 #5

    Mapes

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    Because [itex]-mc(T_f-T_i)=mc(T_i-T_f)[/itex]. Is this what you mean?
     
  7. Nov 17, 2009 #6
    so we should write heat lost by copper as negative since the heat is released out, but the result of [itex]-mc(T_f-T_i)[/itex] will still be positive right?
     
  8. Nov 17, 2009 #7

    Mapes

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    If you're saying that we should include the minus sign prefactor because we're considering the amount of energy leaving, but that the sign of [itex]-mc(T_f-T_i)[/itex] is positive because [itex]T_f<T_i[/itex], I agree.
     
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