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Heat balance problem

  1. Dec 15, 2005 #1
    Here is the problem. 50g of liquid is heated by 30 degrees in a calorimeter in 530 seconds. The power is 10W. 100g is heated by the same amount in the same time with a power input of 16.1W. What is the s.h.c (Cs)of the liquid.

    I know that:

    Cs = (1/mass) x (dH/dt for liquid) + (dH/dt for calorimeter)

    and thus I have a set of two simultaneous equations which can be solved for Cs.

    I also know (well, I think I do) that dH/dt for the calorimeter is equal to the power input. The questions is, what is dH for the liquid? Am I barking up the right tree in the first place?
     
  2. jcsd
  3. Dec 15, 2005 #2

    Astronuc

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    Staff: Mentor

    Well, there are two equations, and also two variables, the specific heat of the liquid and the heat (energy) delivered to the calorimeter, which does have some mass.

    Now one knows that the heating takes place of the same time 530 s. In one case the heat rate is 10 W (or 10 J/s), and in the other case the heat rate is 16.1 W.

    Now set up the two energy balance equations, and solve for the specific heat of the liquid and the energy into the calorimeter.
     
  4. Dec 17, 2005 #3
    Astronuc, many thanks for your help there.

    I read your message yesterday but, as I reached the end of the post, the telephone rang and I was called away with some bad news. Thankfully, things didn't turn out anywhere near as bad as first thought.

    I shall return to the problem over the weekend when things settle down and post my solution up here.
     
  5. Dec 19, 2005 #4
    50g of liquid is heated by 30 K in a calorimeter in 530 seconds. The power is 10W. 100g is heated by the same amount in the same time with a power input of 16.1W. What is the s.h.c (Cs)of the liquid.

    Made a mistake in original post - what I know is that:

    Cs = (1/mass) x [dH/dT]liquid + [dH/dT]calorimeter

    and dH for the calorimeter is IVt (power x time), So:

    Cs(1) = 1/0.05 x [dH/30] + [10 x 530 / 30]
    Cs(2) = 1/0.1 x [dH/30] + [16.1 x 530 / 30]

    Cs(1) = 20[dH/30] + 176.67
    Cs(2) = 10[dH/30] + 284.43

    Rearranging:

    (1) 176.67 = Cs - 20[dH/30]
    (2) 284.43 = Cs – 10[dH/30]

    Subtract (1) from (2):

    107.76 = 10 [dH/30]

    [dH/30] = 107.76 / 10

    dH = 10.776 x 30 = 323.28 J

    Substitute dH into one of the equations:

    Cs = 20 x (323.28 / 30) + 176.67 = 392.19 J Kg-1 K-1

    Does this look right. It seems low for a liquid unless it something like Mercury?
     
    Last edited: Dec 19, 2005
  6. Dec 19, 2005 #5
    My answer to the question:
    Question_2.JPG
     
  7. Dec 19, 2005 #6

    Astronuc

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    Staff: Mentor

    Well, I think you may be making this more complicated than it is.

    I would recommend doing an energy balance and not worry about heat rate (dH/dt), in which case, one has the following general equation:

    (energy into liquid) + (energy into calorimeter) = energy input = power * time

    or mcs[itex]\Delta T[/itex] + Ecal = P * t.

    In the first case one has

    0.05 kg * cs * 30K + Ecal = 10 W * 530 s = 5300 J
    and in the second, one has
    0.10 kg * cs * 30K + Ecal = 16.1 W * 530 s = 8533 J.

    Two equations and two variables. One can multiply the first by (-1) and add to second equation.

    But wait.

    What is the difference between the two cases? The mass of liquid and the power input. All the other parameters are the same.

    The difference in power * time (same in both cases) = difference energy = 8533 J - 5300 J = 3233 J.

    Where does that difference go? Into the difference between the masses (100 g - 50 g) = 50 g or 0.05 kg.

    So the 3233 J = 0.05 kg * cs * 30 K or

    cs = 3233 J / (0.05 kg * 30 K) = 2155.3 J/kg-K.
     
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