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Latent heat

  1. Dec 7, 2005 #1
    I'm not sure if I'm approaching these problems correctly. Could someone check my work?

    Steam at 100°C is condensed in 500g of water at 20°C.

    a. What is the maximum amount of steam that can be condensed in this amount of water?

    Q[steam] = m[steam] * L[c steam] * (T[f] – T)
    Q[steam] = m * 2*10^4 * (20 - 100)
    Q[steam] = m * -1.6*10^6 J

    Q[water] = m[water] * L[v water] * (T[f] – T)
    Q[water] = .5kg * 3.33*10^5 * (100°C – 20°C)
    Q[water] = 1.32*10^7

    Q[steam] + Q[water] = 0

    m * -1.6*10^6 J + 1.32*10^7 = 0

    m = (- 1.32*10^7) / (-1.6*10^6 J)

    m = 8.25kg

    I don't even know what a reasonable answer would be...


    b. If only 5g of steam are condensed, what is the final temperature of the water?

    Q[steam] = m[steam] * c[steam] * (T[f] – T)
    Q[steam] = .005kg * 2*10^4 * (T – 100)
    Q[steam] = 100*T – 1*10^4

    Q[water] = m[water] * c[water] * (T[f] – T)
    Q[water] = .5kg * 3.33*10^5 * (T – 20°C)
    Q[water] = 1.67*10^5 * T – 3.33*10^6

    Q[steam] + Q[water] = 0

    100*T – 1*10^4 + 1.67*10^5 * T – 3.33*10^6 = 0

    100 * T + 1.67*10^5 * T = 1*10^4 + 3.33*10^6

    T (100 + 1.67*10^5) = 1*10^4 + 3.33*10^6

    T = (1*10^4 + 3.33*10^6) / (100 + 1.67*10^5)

    T = 19.99°C


    (I know most of the units are missing-- I'm just lazy... :)

    Thanks,

    dusty.......
     
  2. jcsd
  3. Dec 7, 2005 #2

    Astronuc

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    Staff: Mentor

    There is a mistake in the calculations. The steam condenses from vapor to liquid at the saturation temperature (100°C) at 1 atm (101.325 kPa). There is no temperature change during condensation. The water does increase from 20° to 100°C, which is correct.

    Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c3

    Use heat of vaporization of 2.26 J/kg (539 cal/g) for liquid water to steam, which is the heat given up when condensing from vapor to liquid.
     
  4. Dec 7, 2005 #3
    Thanks Astonuc-- my results seem much more realistic now :)
     
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