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Mass of steam condensed

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) In an espresso coffee machine, steam at 100 °C is passed into milk to heat it. Calculate (i) the energy required to heat 150 g of milk from room temperature (20 °C) to 80 °C, (ii) the mass of steam condensed.

    (b) A student measures the temperature of the hot coffee as it cools. The results are given below:
    d22f1b6d6801.jpg
    A friend suggests that the rate of cooling is exponential. (i) Show quantitatively whether this suggestion is valid. (ii) Estimate the temperature of the coffee after a total of 12 min.
    • Specific heat capacity of milk = 4.0 kJ kg-1 K-1
    • Specific heat capacity of water = 4.2 kJ kg-1 K-1
    • Specific latent heat of steam = 2.2 MJ kg-1
    Answers: (a) (i) 3.6 * 104 J, (ii) 15.8 g

    2. The attempt at a solution
    (a) (i) ΔQ = mcΔθ = 0.15 kg * 4000 J kg-1 K-1 * (80-20 °C) = 36 000 J -- the energy required to heat 150 g of milk from room temperature of 20 degrees to 80 degrees.

    (a) (ii) ΔQ = ml → to find mass m = ΔQ / l = 36 000 J / 2.2 * 106 J kg-1 = 0.016 kg or 16.4 g. Which is wrong. What did I miss?

    (b) (i) Can't say that it's exponential, since the increase is not the same in percentage.

    (b) (ii) The temperature should be 29 °C, because we have a difference of two twice and a difference of one once. So, I guess every difference should be two times the number: 2 x 2, 2 x 1, 2 x 0.

    1b2376b65c9b.jpg

    In sum: what's missing in (a) (ii) and is my logic in (b) (i-ii) correct?
     
    Last edited: Sep 23, 2016
  2. jcsd
  3. Sep 23, 2016 #2

    CWatters

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    Check the temperature change. 80-60 ??
     
  4. Sep 23, 2016 #3
    Oh yes, it's a typo, I guess I was thinking of sixty (80-20 = 60) and that's why got 80-60. The answer is correct though in (a) (i).
     
  5. Sep 23, 2016 #4
    Any ideas please?
     
  6. Sep 23, 2016 #5

    TSny

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    When the steam condenses, what is the temperature of the resultant water immediately after the condensation? What happens subsequently to the temperature of this water as it comes to thermal equilibrium with the milk?
     
  7. Sep 23, 2016 #6
    Sorry, what water are you talking about? I think we only have milk in part (a).
     
  8. Sep 23, 2016 #7

    TSny

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    What happens to the steam when it is added to the milk?
     
  9. Sep 23, 2016 #8
    The steam heats the milk to 80 degrees. It just heats the milk, as I see it.
     
  10. Sep 23, 2016 #9

    TSny

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    The steam is originally at 100 oC when it is added to the milk. At this temperature, what happens to the steam if some heat is transferred from the steam to the milk?
     
  11. Sep 23, 2016 #10
    I would guess it would have some drops or water. The steam would create humidity.
     
  12. Sep 23, 2016 #11
    I think you mean that when we heat milk to 80 degrees using steam, some water is added to milk. That's why we need to calculate the combined mass of water and milk.

    So we have ΔQ = mcΔθ where we need to find the mass of water, m = ΔQ / cΔθ = 36 000 / 4200 * 60 = 0.14 kg is the mass of water.

    Then we combine 16.4 g with 142.9 g = 159.3 g. The mass of steam condensed.
     
  13. Sep 23, 2016 #12

    TSny

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    This is not correct.

    Try the following exercise. How much heat would need to be removed from 100 g of steam at 100 oC to convert it all to water at 90 oC?
     
  14. Sep 23, 2016 #13
    m = 0.1 kg
    TInitial = 100 °C
    TFinal = 90 °C
    TLost = -10 °C
    c of water = 4200 J / kg K

    ΔQ = mcΔθ = 0.1 * 4200 * -10 = -42 J.
     
  15. Sep 23, 2016 #14

    TSny

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    You've correctly calculated the heat required to change the temperature of 100 g of water from 100 oC to 90 oC, except your final number does not have the decimal in the right place.

    But this is only part of the answer.
    How much heat must first be removed from the steam in order for it to condense into water at 100 oC?
     
  16. Sep 23, 2016 #15
    90 - 100 = -10 degrees are lost
    c of steam at 100 degrees is 1890 J / kg K

    ΔQ = 0.1 * 1890 * -10 = -1890 J are lost.
     
  17. Sep 23, 2016 #16

    TSny

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    We have to deal with the steam in two steps:

    (1) The steam condenses to water. This is a phase change. So, think about how to calculate heat involved in a phase change. The temperature doesn't change during the phase change.

    (2) The water, that used to be steam, now cools to the final temperature.
     
  18. Sep 24, 2016 #17
    (1) Heat involved in a phase change: ΔQ = ml = 0.1 * 2.2 * 106 = 220 000 J.

    (2) ΔQ = mcΔ(θNew - θInitial) = 0.1 * 4200 * -10 = -4200 J
    So heat involved 220 000 + (-4200) = 224 200 J

    I'm sorry, but this doesn't get me any closer to understanding what is wrong with my original 16.4 g answer.

    In your example I don't understand whether the mass is constant? 100 g of steam will become 100 g of water?

    I also don't understand what to apply to my problem.

    We have got 36 000 J. Do I need to do the same thing but with c = 4200? Then the answer will be 37 800 J. Using this number I get 17.2 g. If I combine 37 800 + 36 000 I'll get 33.5 g. If I subtract I'll get 1800 J and 0.82 g.

    I also have for some reason 100 °C. If I use Δθ = 100 - 80 or 100 - 20 it still does not get anywhere: 0.15 * 4200 * 20 = 12 600 J or 0.15 * 4200 * 80 = 50 400 J. None of that get the 15.8 g answer.

    If I look for Q = ml = 0.0158 * 2.2 * 106 = 34 760 J. So my Q should be equal to this number. Even if I subract 37 800 - 36 000 I'll get 1800 and then 36 000 - 1800 = 34 200 J and the mass will be m = 15.5 g. Close but still not 15.8 g.
     
    Last edited: Sep 24, 2016
  19. Sep 24, 2016 #18

    TSny

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    OK, good. You've found the total amount of heat given up by 100 g of steam as it goes through the "two-step" process of condensing to water and then cooling to a final temperature of 90 degrees.

    Yes. (Same number of H2O molecules before and after.)

    Yes, that's the amount of heat that must be added to the milk to get the final temperature to be 80 degrees. This heat comes from the steam. That steam will start at 100 degrees and then will go through the "two-step"" process to become water at the final temperature.

    Let M be the unknown amount of steam required. Can you set up an expression for the total amount of heat given up by this amount of steam? Just follow what you did for the example with 100 grams.
     
  20. Sep 24, 2016 #19
    Thank you, this helps.

    Alright, so we have milk that got it's temperature increased from 20 to 80 and the heat required to do that is equal to 36 000 J.

    The steam of mass M was transformed into water of equal mass M and the transaction took ΔQ = ml or Q = (2 200 000 M) Joules. Now we have water that is cooling down from 100 degrees to 80 degrees. ΔQ = mcΔθ or Q = 4200 * (80 - 100) M so Q = -84 000 M.

    We have
    Q = 2 200 000 M and Q = -84 000 M.

    Am I following you correctly?
     
  21. Sep 24, 2016 #20

    TSny

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    OK. But when steam condenses to water it loses heat. So, Q for the steam would be negative: Q = -ML, where Q represents heat added to a system.
     
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