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The heat balance

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    In heat impervious bowl with unknown volume there are unknown amount amount of water(10 degress celcius).
    Metal ball with unknown mass / volume / specific heat(ball temperature was 100 degress celcius) - was landed in water and managed to raise temperature to 40 degress celcius. Then another metal ball with unknown mass / volume / specific heat(ball temperature was 100 degress celcius too) was landed. Question: What was the final temperature of water in bowl?

    2. Relevant equations
    We used Qgave=Qreceived(The heat balance equation)
    Cwater x Mwater = Cball x Mball x 60(temperature difference after second ball) / 30(difference after first land) =
    2CballMball
    So we got CwaterMwater = 2CballMball


    3. The attempt at a solution
    CwaterMwater(40-10) = CballMball(100 - 40)
    CwaterMwater(t-40)
    CballMball(100-t)
    We concidered fact that second landed ball gives the heat not only to the water but to the first bowl also cause
    of temperature drop

    Sorry for grammar mistakes , english isn't my first language
     
  2. jcsd
  3. Nov 15, 2016 #2

    gneill

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    Staff: Mentor

    I assume that the first and second metal balls are identical (material, mass)?

    You are correct that the second ball will be sharing its heat with both the water and the first ball. Since that is the case, can you write a heat balance equation for the steady state (long after the second ball was added)?
     
  4. Nov 15, 2016 #3
    They're identical in all aspects. There in data time wasn't given , so i think not long after , just when first ball and water temperatures levels together into one
     
  5. Nov 15, 2016 #4

    gneill

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    The question asks for the final temperature. So that's the steady state that evolves after sufficient time passes. You need to write an energy balance equation that represents how the heat moved to its final distribution.
     
  6. Nov 15, 2016 #5
    How to do write energy balance equation if Δt is the only i got
     
  7. Nov 15, 2016 #6

    gneill

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    When in doubt, start by writing it out using symbols (variables) to represent the quantities involved.
     
  8. Nov 15, 2016 #7
    Still haven't managed way to calculate this , maybe you can help a bit?
     
  9. Nov 15, 2016 #8

    gneill

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    I can't write the equation for you, you need to make an attempt. But I can give a hint. On the left hand side put the new source of heat where you write the expression for the heat it gives up in going from 100C to some final temperature Tf. On the right hand side you write the expression for heats taken up by the existing materials going from 40C to Tf. Use variables to represent the unknown things like specific heats and masses.
     
  10. Nov 15, 2016 #9
    I managed to get 140+2Tf/60 = Tf-40/30 , how to get final answer?
     
  11. Nov 15, 2016 #10

    gneill

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    I can't validate that expression because I don't know what you did to derive it.
    What is t?
     
  12. Nov 15, 2016 #11
    Ups Tf
     
  13. Nov 15, 2016 #12

    gneill

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    Can you show your work in detail, starting from the energy balance equation for when the second ball is dropped in?
     
  14. Nov 15, 2016 #13
    Okay:

    Findings: (final temperature) t - ?
    Data: t1 - 10dc(Just waters temp)
    t2 - 100dc(First ball)
    t3 - 40dc (After first ball)
    Work:
    // We create variables c and mass for equation
    CbMb(t2-t1) = CwMw(t3-t1) // water temperature becomes equal to ball because of balance
    CbMb(100-40) = CwMw(40-10) // i just inserted numbers.

    // While we land second ball the heat energy goes to the ball and to the water soo...
    CbMb(t2 - t) = CbMc(t3-t) + CwMw(t3-t) // All three temperatures equals and we can also say:
    CbMb(100-t) - CbMb(t-40) = CwMw(t-40) // If you subtract the heat the first ball takes you get the temperature only water gets
    // Now we get the first amounts of heat ball gave and water received and devide the differece of balls to that equation
    CbMb(100-t) - CbMb(t-40) = CwMw(t-40) / CbMb(100-40) = CwMw(40-10)
    When we got this done with all numbers we remove variables
    100+40+2t / 60(got this from 100-40) = t-40 / 30(got this from 40-10)

    At the finale i got 140-2t / 60 = t-40 / 30 and don't know whats more
     
    Last edited: Nov 15, 2016
  15. Nov 15, 2016 #14

    gneill

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    The first term should be CbMb(t2-t3). Note that it's okay to just use the numbers for the temperatures since they are known values here. So your next line is what I would have expected:
    Good. So now you know a relationship between CwMw and CbMb, right? What is

    ##\frac{CwMw}{CbMb} = ?##

    Okay, you have to be careful to get the ordering of the temperatures right so that the signs of the terms will be right. Otherwise you'll end up gaining or losing energy when you should be doing the opposite.

    In the expression above the left had side will be positive and represents the energy being "given" by the second ball. We expect t2 to be greater than the final temperature so that t2 - t will be positive. Good.

    Now on the right hand side we need to sum up the energies that are received by the other materials. We want these terms to be positive too. But look at the ΔT you've written: t3 - t. t3 is 40C and we expect t to be greater than that. So that would turn both terms on that side of the equation negative. So you need to reverse those variables. Then you'll have:

    CbMb(100 - t) = CbMb(t - 40) + CwMw(t - 40)

    Now I suggest that you find a way to get rid of the CbMb and CwMw variable groups. Recall from above that you found a relationship between CbMb and CwMw? Suppose you were to divide through the whole equation by CbMb, what would you be left with? Can you make all those unknown parameters disappear?
    I don't understand what you were trying to do in this last part. Perhaps because I don't understand the equation you've written with two equals signs. However, if you can do as I suggested above (clearing away the unknown parameters) then I think you'll be on a more straightforward path to the answer.
     
  16. Nov 15, 2016 #15
    But there is no unknown parameters left , there are just a numbers and final temperature what i need to find

    somehow i need to subtract or devide these numbers to get only temperature

    140 - 2t / 60 = t-40 / 30 ( no unknown left)
     
  17. Nov 15, 2016 #16

    gneill

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    I don't understand your method for getting those numbers, and from what I can tell (making assumptions about the order of operations to interpret your expression), it cannot be correct because the t cancels on both sides. So the equation cannot be correct.

    Inserting parentheses to make clear the order of operations I interpreted your equation as:

    (140+2t) / 60 =( t-40) / 30

    Reducing the LHS to have a denominator of 30:

    (70 + t)/30 = (t - 40)/30

    so that:

    70 + t = t - 40

    which is impossible.
    EDIT: The starting expression was later corrected by the OP. So redoing the math:

    (140 - 2t) / 60 =( t - 40) / 30 (note the "-" on the LHS)
    (70 - t)/30 = (t - 40)/30
    70 - t = t - 40
    2t = 110
    t = 55
     
    Last edited: Nov 15, 2016
  18. Nov 15, 2016 #17

    gneill

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    The result can also be obtained from

    CbMb(100 - t) = CbMb(t - 40) + CwMw(t - 40)

    by dividing through by CbMb and using the previously found value for (CwMw)/(CbMb).
     
  19. Nov 15, 2016 #18
    Thank you once again for the time.
     
  20. Nov 16, 2016 #19

    PeroK

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    Two ideas to make this simpler:

    1) Define ##h_b = C_bm_b## and ##h_w = C_wm_w##, then these represent the amount of heat energy required to raise the water and the ball by ##1K##.

    The first experiment tells you that ##30h_w = 60h_b##, hence ##h_w = 2h_b##.

    The two balls combined, therefore, require the same energy as the water to change by ##1K##.

    2) The final tempreature is the same whether you put both balls in together or one at a time.

    If you put them in together, then, as ##h_w = 2h_b##, the final temperature is the average of the starting temperatures. I.e. ##55 °C##
     
    Last edited: Nov 16, 2016
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