Need some clarification for a latent heat related problem

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Byrgg
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Ok, I don't really need to know because it's something I've been told to ignore in the text, but I'm curious about it anyway, here goes:

At a nuclear power plant, water at 8 deg. celsius from a nearby lake is used in a heat exchanger to condense spent steam, at a temperature of 120 deg. celsius, to water, at 85 deg. celsius before it is recycled into the reactor. If the cooling water returns to the lake at a temperature of 19.deg celsius, how many kg of water are needed per kg of steam? Ignore the pressure changes on the steam.

Here's how I solved it, could you tell me if I'm right?

Q_gained(1) = -Q_lost(2)

Q_1 = mcT (for the water)
Q_2 = mcT + ml + mcT (for the steam)

specific heat capacity of water is 4200
specific heat capacity of steam is 2000
latent heat of vapourization of water is 2255000

mcT = -(mcT + ml + mcT)

m(4200)(19-8) = -(m(2000)(100-120) + m(2255000) + m(2000)(85-100))

m46200 = -(m(-40000) + m(2255000) + m(-3000))
m46200 = m4000 - m2255000 + m3000

divide both sides by m2... (mass of the steam)

m1(46200)/m2 = 4000 - 2255000 + 3000

divide both sides by 462000...

m1/m2 = (4000 - 2255000 + 3000)/46200
m1/m2 = 48.66

Therefore the ratio of water to steam is 48.66, the book says 50, if I followed the significant figures than I would've acheived 50 I'm pretty sure so, I'm confident I did that right does it look ok? But now here's what I want to know:

I want to know exactly how the pressure changes in the steam would affect the answer(considering it told me to ignore them).

Thanks in advance, someone please respond soon :smile:
 
on Phys.org
Byrgg said:
Someone please help.
As for the calculations they seem to be right, as you have used the correct formulas to solve it.Your answer may be a more accurate answer as oppose to the one listed in the book. The pressure of the steam would differ the result of the calculation because a higher pressure would result in there being more steam in a set volume, than the same volume measured at a lower pressure. Therefore, since there is more steam in an environment with a higher pressure, less kinetic energy needs to be removed before the steam turns back to water because the pressure has raised the boiling point of water, i.e, the water now boils at a higher temperature.
Pavadrin
 
Ok thanks for the help. So basically, the I would have to use a different boiling point if I didn't ignore the pressure changes in the steam?
 
Byrgg said:
Ok thanks for the help. So basically, the I would have to use a different boiling point if I didn't ignore the pressure changes in the steam?
Correct.
Pavadrin
 
Oh, also, increasing the pressure, increases the boiling point? Because in a high pressure system, the substance is already closer to reaching a liquid(since liquids are more dense than their gas counterpart)? Did I get that right?
 
Yes increasing the pressure increases the boiling point, as the pressure is the external force which is pushing the steam particles together. When a substance boils, the particles have gained enough kinetic energy to break free from the external pressure. An example of this is seen when water boils at sea level at 100 degrees Celsius, however mountaineers at the top of Mt. Everest experience the water boil at approximately 70 degrees Celsius, as there is less atmospheric pressure at higher altitudes.
Pavadrin.
 
Just further to what pavardrin said, a substance will boil when there some of the molecules in a substance have enough thermal (kinetic) energy to overcome the intermolecular forces within the substance. A useful definition of boiling point is, "the temperature at which the vapor pressure of the liquid equals the pressure of the surroundings." which is basically what pavadrin said above.