Latent heat of vaporization of H2O ?

[mr.coon]

Suppose the latent heat of vaporization of H2O were twice its actual value.

(a) Other things being equal, would it take the same time, a shorter time, or a longer time for a pot of water on a stove to boil away?

(b) Would the evaporative cooling mechanism of the human body be as effective, less effective, or more effective?


i need a bone thrown to me on this one.

attempt at (a):

latent heat of vaporization on water = 22.6e5so twice the actual value would be 4.52e6. to me it would take the same amount of time for the water to boil out because the same about of steam would be produced no matter the boiling point.

attempt at (b):

i have no clue.

 

[Borek]

In both questins you have to realize that you need twice amount of energy to evaporate the same amount of water.

(a) Why does the water on the stove boil?

(b) How does the colloing mechanism work? What happens to sweat?

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[mr.coon]

(a) water boils because it is experiencing rapid evaporation and turning to a gas state. the bubbles are the gas escaping.

i realize that it would take twice the energy to boil the water but what does that have to do with the rate at which it evaporates? i am beginning to lean towards it taking longer to evaporate the water simply because more work is involved and we have constant atmospheric pressure.

(b) sweat evaporates off the skin cooling the body. so in this case it would take twice as much energy/heat to evaporate the sweat thus making the cooling effect less effective.

 

[Borek]

(a) water boils because it is experiencing rapid evaporation and turning to a gas state. the bubbles are the gas escaping.

i realize that it would take twice the energy to boil the water but what does that have to do with the rate at which it evaporates? i am beginning to lean towards it taking longer to evaporate the water simply because more work is involved and we have constant atmospheric pressure.

Think: if you turn the stove up, will the boiling speed change?

(b) sweat evaporates off the skin cooling the body. so in this case it would take twice as much energy/heat to evaporate the sweat thus making the cooling effect less effective.

So, 1 g of sweat will remove twice the amount of heat from your body, but the cooling will be twice less effective?

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methods

 

[mr.coon]

Borek, LOL i know you are trying to pull me through the tunnel but i am having a hard time fitting.

(a) the higher you put a stove the faster the water boils and the more rapidly the water evaporates out. if you boil two pots of water at the same speed and one has twice the boiling point they would take the same amount of time to evaporate out to me regardless if one needs more heat to boil because they are still boiling at the same speed. i am sure this is wrong though because it was what i thought in the first place and the answer is that it will take less time. which if that is the case it is also confusing because i know for a fact that if the boiling point of water is reduced to a 10th of its natural point it would less time to boil out than a regular pot of regular water. i don't know the explanation of that, but it read it elswere in a physics publication while trying to figure out this question.

(b) ok, so the cooling system would be twice as effective because the water will take away twice as much heat from the body as it evaporates. what keeps popping into my head is that the body would then have to get twice as hot to make the sweat evaporate? i guess i need to forget about that. as above, i already know that the evaporative cooling mechanism of the body would be less effective if water evaporated at a 10th of its natural value. It would
only take one 10th the amount of heat away from the body to evaporate water from the skin reducing the cooling effect.

 

[Borek]

(a) the higher you put a stove the faster the water boils and the more rapidly the water evaporates out.

Good. The question is - why?

if you boil two pots of water at the same speed and one has twice the boiling point they would take the same amount of time to evaporate out to me regardless if one needs more heat to boil because they are still boiling at the same speed.[q/uote]

Two pots of water have the same boiling point. However, to evaporate some amount of water you need to deliver enough energy. How does the speed at which the energy is delivered depend on the stove setting?

i know for a fact that if the boiling point of water is reduced to a 10th of its natural point it would less time to boil out than a regular pot of regular water. i don't know the explanation of that, but it read it elswere in a physics publication while trying to figure out this question.

You must be misreading it. Boiling point doesn't matter - as long as water is already at the boiling point speed of evaporation depends only on amount of heat delivered in time (that is - heating power). Could be they meant time to get water to boiling is shorter.

(b) ok, so the cooling system would be twice as effective because the water will take away twice as much heat from the body as it evaporates.

Exactly. That means that to cool down you would need to perspire twice less.

what keeps popping into my head is that the body would then have to get twice as hot to make the sweat evaporate?

No, you are obviously mixing two separate things. Boiling point doesn't matter. To evaporate water you have to deliver heat. You have to deliver heat to evaporate water even if you are far from the boiling point. Basically that's the same amount of heat (although evaporation heat is to some small extent temperature dependent, so it is in reality not EXACTLY the same amount of heat).

i already know that the evaporative cooling mechanism of the body would be less effective if water evaporated at a 10th of its natural value. It would
only take one 10th the amount of heat away from the body to evaporate water from the skin reducing the cooling effect.

Right.

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[mr.coon]

(a) sorry, i misquoted the reference. it said "if latent heat of vaporization" cut to a 10th of its original value. not "boiling point". here is the full quote, "If the latent heat of vaporization Lv of H2O were one tenth of its actual value. Other things being equal, if the stove is supplying heat at a constant rate it would take less time for the water to boil out.

ok, back to our case where the LHOV is doubled. the boiling point is the same for the two pots. the latent heat of vaporization is different from the boiling point. so that means they both would be boiling at the same temp however the doubled one would take longer to boil out because it needs more energy to vaporize the water.

i have been thinking the boiling point and latent heat of vaporization were the same thing.

 

[Borek]

i have thinking the boiling point and latent heat of vaporization were the same thing.

It was becoming more and more visible

I suppose now answers to both questions are obvious?

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[mr.coon]

thanks for sticking it out with me. my textbook sucks. it had the two discussed side by side and never differentiated between the two. i just read another article where it explained this.
it seems obvious now.

final try:

(a) would take a longer time to boil away

(b) the cooling system would be twice as effective

 

[Borek]

hi5

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[mr.coon]

alright! you learn something new every day!