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Thermodynamics ice and water bath

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    a bath contains 1000 grams of water and 100 grams of ice in thermal equilibrium. at some point in time a 500gram block of stone(granite) at 100 deg celsius is added to the bath

    a) what is the temp of the bath before the block of stone is added

    b) what is the temp of the bath and stone when a new equilibrium is reached?

    c) how much did the entropy of the ice increase as it melted?


    2. Relevant equations

    temp equil = (mi Ci Ti + mw Cw Tw)/((mi Ci) + (mw Cw)) where
    mi = mass of ice, mw = mass of water, Ci - specific heat of ice, Cw is specific heat of water, Tw = temp of water --not given, Ti = temp of ice -- not given

    i think i use a similar equation for part b, except i THINK i average the specific heats of ice and water

    so for part b)

    temp equil = (mb Cb Tb + ms Cs Ts)/((mb Cb) + (ms Cs)) where
    mb = mass of stone, Cb = specific heat of stone, granite 1.2, Tb = temp of stone, ms = mass of ice and water, Cs = avg of specific heat of ice and water, Ts = equil temp from (a)

    3. The attempt at a solution

    i tried the above equations but didn't get far because the original temp of the ice and water separately was not given, the solution for part a is supposed to be 0 deg celsius but i did not get it

    as for part b, i am not sure if i am even supposed to average the specific heats, when after finding out that the answer for part a was 0, i put in the Ts as 273K and used the other givens for the block and got 11.5 deg celsius -- the actual answer is supposed to be 5.13 deg celsius

    i should be able to figure out part c after finding out how to get 0 for a, and 5.13 for b.

    is my approach completely off? any help appreciated
     
  2. jcsd
  3. Jun 1, 2008 #2

    alphysicist

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    Hi scholio,

    For part a, the important point is that the ice and water are in thermal equilibrium. What does being in thermal equilibrium mean? Once you answer that, I think you will see why the given answer is correct.

    For part b, I don't think you did it correctly. You don't want to average the specific heats, just write down separate terms for the water and the ice. Also, did you calculate the heat required to melt the ice? A good starting point for this problem would be:

    (heat gained by ice and water) = (heat lost by granite)

    and you'll have several terms on the left hand side. What do you get?
     
  4. Jun 1, 2008 #3
    alphysicist, i understand that in order for the water and ice to exist together in the bath the temp must be atleast 0 deg celsius, but how do go about getting that number by calculating it out of curiosity?

    so are you saying for part be to use mCdeltaT (water-ice)= mCdeltaT (stone), if so what is the specific heat for the water-ice, i came upon that issue and thus tried the whole averaging thing.

    thanks again
     
  5. Jun 1, 2008 #4

    alphysicist

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    You don't calculate it; they are giving it to you. The question is testing whether you know what thermal equilibrium means and whether you know what the limits on the temperatures of ice and water will be under "normal" conditions.

    There's no need to have just one term on the left hand side. The left hand side has three things happening: the ice is melting, the 1000 grams of water is heating, and the 100 grams of water (that used to be ice) is heating. My preference is to have three terms on the left hand side, but of course you could combine the two heating terms (since they are both just water).

    But more importantly, what formula gives the heat required to melt ice? It can't be of the form m C (delta T), because the temperature does not change while the ice is melting.

    Once you get that, the equation is:

    (heat to melt ice) + (heat to raise temp of 1000 g water) +(heat to raise temp of 100 grams water) = (heat lost by granite cooling)
     
  6. Jun 2, 2008 #5
    i am still having trouble, i am getting -23 deg c, not the 5.13 deg i am supposed to get.
    this is what ive done

    (heat to melt ice) + (heat to raise temp of 1000 g water) +(heat to raise temp of 100 grams water) = (heat lost by granite cooling)

    so

    miLf + mwCwdeltaTw = msCsdeltaTs

    solve for T, the equilibrium temperature of the water bath and stone
    where
    mi = 100grams ice = 0.1kg,
    mw = 1100 grams of water --> orginal water and water from melted ice = 1.1 kg,
    Cw = specific heat of water = 4.186,
    ms = mass of stone = 500grams = 0.5kg,
    Cs = specific heat of stone = 1.200
    Ts = temperature of stone = 100deg celsius = 373 kelvin
    Tw = temp of water, equilib temp so 0 deg celsius --> 273kelvin
    Lf = latent heat of fusion, water = 333kj/kg

    one again i am getting 4.0T = 1000, so T = 250k = -23 deg celsius, i'm supposed to get 5.13 deg celsius


    help appreciated thanks
     
    Last edited: Jun 2, 2008
  7. Jun 2, 2008 #6

    alphysicist

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    Hi scholio,

    I got 5.13 degree celsius, so let me look at your numbers.

    I notice you have the masses in kilograms, but the specific heats in J/(gram degree). I would leave the masses in kilograms, and then the specific heat values would be 4186 for water and 1200 for the stone.

    It's not an error, but you can leave the temperatures in celsius here. The temperature only appears in this equation as a difference in temperature, and the change in temperature is the same in celsius or kelvin.

    So far I'm not able to see how you got -23 degrees as the answer. So try changing the specific heats from above to be consistent with units of kilograms. If you still don't get the answer, could you post the numbers you used and the results for the three separate terms of this equation:

    If I had to guess, I would say that it's likely you are not putting in the equilibrium temperature T in the equation correctly, but I can't know for sure until you post what those three terms are.
     
  8. Jun 2, 2008 #7
    Tw = T - T_w where T_w is 273K equil temp of water/ice
    Ts = T - T_s where T_s is temp of stone before entering bath, 100 deg cels

    miLf + mwCwdeltaTw = msCsdeltaTs
    0.1(333) + (1.1)(4.186)(T - 0) = 0.5(1.200)(T-100)
    33.3 + 4.60T - 0 = 0.6T - 100(0.6)
    4.0T = -60 - 33.3
    4.0T = -93.3
    T = - 23.3 celsius
     
    Last edited: Jun 2, 2008
  9. Jun 2, 2008 #8

    alphysicist

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    You have a sign error on the right side of your equation. There are two ways to think about it:

    1. When we use:

    (energy gained) = (energy lost)

    we mean the magnitudes, so the energy lost would be given by

    0.5(1.200)(100-T)

    (That's how much energy is leaving the stone; we want the magnitude, so we subtract the larger from the smaller.)



    2. If you don't want to use magnitudes, and you would like for [itex]\Delta T[/itex] to always mean T_f - T_i, then you would start with:

    (energy change of ice+water) + (energy change of stone) = 0

    (energy change of ice+water ) = - ( energy change of stone)

    (energy change of ice+water ) = - (ms Cs (100 - T) )

    so the right hand side still becomes (ms Cs (T - 100) )
     
  10. Jun 2, 2008 #9
    aphysicist, thanks again, i made the change you suggested and got the answer i was looking for, 5.13 deg celsius

    cheers

    as for part b, i need to find how the entropy increased as the ice melted, this is what i did, i'm supposed to get 0.122kj/k

    deltaS = deltaQ/T = miCideltaTi/Ti = ((0.1)(2.093)(5.13))/(5.13) = 0.20

    did i interpret the problem incorrectly?
     
    Last edited: Jun 2, 2008
  11. Jun 2, 2008 #10

    alphysicist

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    Not quite. Here they want the entropy change for just the ice melting. How much heat was required to melt the ice? What temperature (now it has to be in Kelvin, since it's not a temperature change) is the process occurring at? Those are the questions to answer, and those answers will give you the entropy change.
     
  12. Jun 2, 2008 #11
    i think i got it:

    first use the heat needed to melt 100g of ice using Q = mLf = 0.1(333) = 33.3 joules, then divide that my temperature which we found to be 5.13 deg celsius = 278.13 K

    so entropy = 33.3/278.13 = 0.1197 = 0.12 joules/kelvin

    is that close enough, i'm supposed to 0.122
     
  13. Jun 3, 2008 #12

    alphysicist

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    Almost, but the ice doesn't melt at a temperature of 278.13 K. Once you get the right temperature you'll get the answer.
     
  14. Jun 3, 2008 #13
    you're right, ice melts at 0 deg celsius aka 273 K, subbing that in for T i got 0.1219 = 0.122 !
     
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