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Specific Heat Problem

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I have some boiling water that I am adding potatoes to. I am asked to find the final temp after the potatoes have been added to the boiling water.

    Tiw=100C (initial boiling water temp)
    Tip=30C (initial room temp potatoes)
    mp=0.5kg (mass of potatoes)
    mw=1.5L or 1.5kg (mass of water)
    cw=4.2kj/kg*C (specific heat of water)
    cp=3.4kj/kg*C (specific heat of potatoes)

    2. Relevant equations

    [itex]C=\frac{Q}{\Delta T}[/itex]
    [itex]c=\frac{C}{m}[/itex]

    Since no heat is assumed to be lost, Qlost=Qgained

    Subscript w is for water and p is potato

    3. The attempt at a solution

    Qlost=Qgained

    cwmw(Tf-Tiw)=cpmp(Tf-Tip)

    Both objects reach thermal equilibrium therefore Tf is the same.

    After some algebra I end up with:

    [itex]T_f = \frac{(c_w m_w T_iw) - (c_p m_p T_ip)}{(c_w m_w) - (c_p m_p)}[/itex]

    Then plugging in my given values I end up with 125C which is no way correct.

    I am thinking my error might be how I am choosing my Ti/Tf values. I saw an example somewhere which had a couple of the values switched. Which didn't really make sense because as far as I know, it's always Tf-Ti?

    Where am I making my error? If needed I can post my algebra steps. I did them a few times and got the same answer every time...
     
    Last edited: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2
    However if I swap the temp values like an example I saw....

    cwmw(Twi-Tf)=cpmp(Tf-Tpi)

    After the algebra I end up with...

    [itex]T_f = \frac{(c_p m_p T_{pi}) + (c_w m_w T_{wi})}{(c_w m_w) + (c_p m_p)}[/itex]

    I get 85.125C. Which is a lot more reasonable.


    If that's the case, why do the temps get represented like that?
     
  4. Jan 21, 2013 #3

    ehild

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    Tf-Tw<0 and TF-Tp>0. If you say that the water loses a negative amount of heat, it means it gains heat.

    The water loses cwmw(Tw-Tf) heat. The potato gains cpmp(Tf-Tp) heat, and they are equal.

    Or you can say that the heat transferred to a substance is cm(Tf-Ti) which can be either positive or negative and the sum of the transferred amounts of heat is zero, as heat is not lost.

    ehild
     
  5. Jan 21, 2013 #4
    So in this case ΔT isn't necessarily Tf-Ti like most Δ's. It's more for interpretation...if that makes sense.

    I get what you are saying. For this problems the final and initial are "swapped" since the water has to lose heat at the end. If I do final - initial, the ΔT is negative and therefore gains heat, which isn't the case.

    Therefore I need to look at what ΔT is doing and set it up from there? Or am I just reaching for a reason why they are swapped and this doesn't make sense?

    Thanks for the reply!
     
  6. Jan 22, 2013 #5

    SteamKing

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    Remember the Zeroth Law of Thermodynamics:

    Heat flows from a body with a higher temperature to a body with a lower temperature.
     
  7. Jan 22, 2013 #6

    ehild

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    You need not think if you add up the terms cm(tf-ti) and make the sum equal to zero.


    ehild
     
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