Latent heat

  • Thread starter dustybray
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  • #1
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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.......
 

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
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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.
 
  • #3
10
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Thanks Astonuc-- my results seem much more realistic now :)
 

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