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Calorimetry problem

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data
    A glass cup with a mass of 0.1 kg and an initial temperature of 23◦C is filled with 0.3 kgof water at 80◦C

    A)What is the final temperature of the water and cup?
    B)How much heat must be added to raise the temperature to 90◦C?
    C) What is the change in the internal energy of the water?
    2. Relevant equations
    E = Q - W
    Q = mcdeltaT

    3. The attempt at a solution
    A) heat gained by glass = heat lost by water

    [itex] m_{glass}c_{glass}(T_f - 23 C) = m_{water}c_{water} (80 C - T_f) [/itex]

    where cglass = specific heat of glass = 840 J/kgC
    cwater = spec heat of water = 4186 J/kgC
    and Tf is the final temp

    [itex] (.1 kg) (840 J/kgC) (T_f - 23 C) = (.3 kg)(4186 J/kgC)(80 C - T_f) [/itex]

    [itex] 84T_f - 1932 = 100464 - 1256T_f [/itex]

    [itex] 1340T_f = 102396 [/itex]

    [itex] T_f = 76.4 C [/itex]

    B) Q = heat gained by glass + heat gained by water

    [itex] Q = m_{glass}c_{glass}(T_f - 76.4) + m_{water}c_{water}(T_f - 76.4) [/itex]

    [itex] Q = (T_f - 76.4)(m_{glass}c_{glass} + m_{water}c_{water})[/itex]

    [itex] Q = (90 - 76.4)(.1(840)+ .3(4186))[/itex]

    [itex] Q = (13.6)(84 + 1256)[/itex]

    [itex] Q = 18221 J [/itex]

    C) since no work is done, the internal energy of the water would be the same as Q for the water
    [itex] E = Q = m_{water}c_{water}(80 C - 76.4 C) [/itex]

    [itex] E = Q = .3(4186)(80 C - 76.4 C) [/itex]

    [itex] E = Q = .3(4186)(3.6) [/itex]

    [itex] E = Q = 4520 J[/itex]

    is this the correct solution?
     
  2. jcsd
  3. Oct 31, 2014 #2

    ehild

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    Gold Member

    The question is the change of internal energy, ΔE, which is equal to the added heat . It is not the same as the internal energy E.
    But ΔE has sign, you have to indicate.
     
  4. Nov 1, 2014 #3
    the water loses heat to reach the final temp so it would be negative?
     
  5. Nov 1, 2014 #4

    ehild

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    Yes.
     
  6. Nov 1, 2014 #5
    Thank you!
     
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