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Heat Transfer and capacity

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Is my approach correct? Thanks all :)

    a)The specific heat capacity of water is 4200Jkg-1K-1 while the specific latent heat of steam is 2.26x106Jkg-1. Explain what is meant by the statement in italics.

    b) Explain, in terms of molecular motion, the large difference between the two values.

    c) A polystyrene cup contains 80g of water at 20C. steam at 100C is passed into the water until the final temperature of the water rises to 80C. Determine the mass of steam Which has been passed into water.

    d) The heat transfer to the cup may be neglected in the calculation. Why?

    2. Relevant equations


    3. The attempt at a solution

    a) Specific heat capacity is the heat energy required to raise the temperature of a material weighing 1kg by 1 degree Celsius.

    The specific latent heat of steam is the heat required to change 1 kg of water to steam without a change in temperature.

    b) In terms of molecular motion it will require much more heat energy for the kinetic energy of the molecules, to reach a certain point where no force of attraction is present. Therefore this change of state requires more energy, rather than the energy required to heat up the liquid.

    c) Q=mc∆θ
    Q=0.080*4200*60
    Q=20160J Amount of heat energy given from the steam to the water.

    Q=ml
    20160J=m*2.26x106Jkg-1
    m=8.92x10-3kg Mass of steam passed into the water.

    d) Polystyrene is a poor conductor of heat unlike metal, this material will act as a good insulator thus assuming heat losses are negligible.
     
  2. jcsd
  3. Apr 23, 2015 #2

    BvU

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    Re c): the latent heat, as you stated, is the energy to condense the steam of 100 C to water of 100 C. So not to water of 80 C.

    Under a) I would perhaps also say something about pressure
     
  4. Apr 23, 2015 #3
    Oh ok, that's a better understanding of the question c. Are the calculations correct?
     
  5. Apr 23, 2015 #4
    according to wikipedia
    So why pressure for a) ?
     
  6. Apr 23, 2015 #5

    BvU

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    Therefore: no, not entirely.
     
  7. Apr 23, 2015 #6
    Heat Lost = Heat Gained
    mc∆θ=mc∆θ
    m*2.26x106Jkg-1*20=0.08*4200*60
    m=4.46x10-4kg

    Is this correct?
     
  8. Apr 23, 2015 #7

    BvU

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    What is it ? 1 * 20 , 60 ?
     
  9. Apr 23, 2015 #8
    Sorry I arranged the units.

    Heat Lost = Heat Gained
    mc∆θ=mc∆θ
    m*2.26x106Jkg-1*20C=0.08kg*4200*60C
    m=4.46x10-4kg

    20 and 60 are the change in temperature.

    Is this correct?
     
  10. Apr 23, 2015 #9

    BvU

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    Because the given value is at 1 Bar. At 0 C it is 2500, at 300 C it's a mere 1400 kJ/kg.

    But I grant you that you could also specify by stating the temperature instead of the pressure :smile: .

    And Boa won't have a lower score if this kind of detail isn't added, I suppose.
     
  11. Apr 23, 2015 #10

    BvU

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    I shouldn't have asked about the 60. I understand the 1 is a typo from -1. and the 20 is a ##\Delta##T.

    But if you look: the left hand side isn't an energy any more if you multiply with 20 degrees.
     
  12. Apr 23, 2015 #11
    Oh right so I should just re-write it like this?

    m*lf=mc∆θ

    lf being the latent heat of fusion.
     
  13. Apr 23, 2015 #12
    But that would be your post 1 attempt and that is wrong a bit.
     
  14. Apr 23, 2015 #13

    BvU

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    You want to write a decent heat balance. Start by distinguishing mass of steam and mass of water instead of using one single symbol m for both. Heat lost = heat gained is a good continuation (you have advance knowledge who gives off and who gains, otherwise the next step would be something like energy before = energy afterwards).

    You already had ##m_{\rm steam} \; l_f ## on the heat lost side. That gives you ##m_{\rm steam}## kilograms of water at 100 C.
     
  15. Apr 23, 2015 #14
    Like this? Sorry if I might be frustrating you, I don't blame you.

    Left side is the water added to the cup, assuming the water is at 100C.

    mc∆θ=mc∆θ
    m*4200(100-80)=0.08*4200*(60)
    m=0.24kg
     
  16. Apr 23, 2015 #15

    BvU

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    Ok, but now the heat of condensation ( = latent heat of vaporization) has disappeared !
     
  17. Apr 23, 2015 #16
    Oh right.

    mc∆θ=mc∆θ
    m*4200*2.26x106Jkg-1*(100-80)=0.08*4200*(60)
    m=1.06x10-7kg

    Q=ml is the energy used to convert the water into steam and vice versa without a change in temperature, and the specific heat capacity * change in temperature indicates the heat lost?
     
  18. Apr 23, 2015 #17

    BvU

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    It is a lot simpler: 1 kg of steam at 100 C condenses to water of 100 C, releasing 2260 kJ of energy. Then that 1 kg of water at 100 C releases another 20 * 4.2 kJ to cool down to 80 C. Together that's ##l_f + Cp \Delta T\ \ ##kJ/kg of steam, way too much for the poor little 80 grams of water at 20 C. As you see, a small correction to the 2260 kJ/kg from the latent heat only, but not to be ignored.

    Now you should have a good idea how much steam is needed in your exercise...
     
  19. Apr 23, 2015 #18
    I'm confused, now I get the idea that its written like this?

    (ml+cΔT)=mc∆θ
     
  20. Apr 23, 2015 #19

    BvU

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    Almost: $$m_{\rm steam} \left ( l_f + c_p \Delta T_1 \right ) = m_{\rm water} c_p \Delta T_2$$
     
  21. Apr 23, 2015 #20
    Oh alright this makes more sense now.

    So the mass is:8.6x10-3kg
     
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