You can evaporate water without boiling it (that’s how clothes get dry!). It will use only latent heat in order to change the phase of the system and won’t heat up water on the process.

How to estimate the energy consumed to evaporate the water without boiling it? How many kJ will be consumed (from the environment, latent heat) by the evaporation of 1 liter of water contained in clothing washed at 40°C put on a drying rack?

According to this page: https://en.wikipedia.org/wiki/Latent_heat, reference 10, I could estimate that, if the clothing are at 35°C, it would take 2500-2,36*35+0,0016*35*35-0,00006*35*35*35j/g, which is 2416j to evaporate 1g of water. So this would give us 2416kJ for 1 liter of water, so approximately 671kWh (according to google converter).

It's a poor computation and I'm surprised we raise almost 1kWh, which is a lot (I bet some electric dryer may consume less to evaporate 1 liter of water from some clothing...)

Would that be correct?

  • $\begingroup$ Are you familiar with Hess' law? $\endgroup$ – Chet Miller May 2 '17 at 12:13
  • $\begingroup$ @ChesterMiller: Not at all $\endgroup$ – jpo38 May 2 '17 at 12:20
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    $\begingroup$ Please Google Hess' law. It will give you the information on how to proceed to get the heat of vaporization at one temperature, knowing the heat of vaporization at another temperature and the heat capacities of the liquid and vapor. $\endgroup$ – Chet Miller May 2 '17 at 12:42

You should be able to find latent heat of evaporation for that temperature.

The vapour will take the energy for vaporization with it just like it would if it were boiling.

If you can find the heat taken from surroundings for vaporization at 100°C, doing it at 40°C would require adjusting what latent heat of evaporation you use.

I was going to write a quick comment to this effect; but it felt more like an answer. I'll leave the solution with the numbers for you though.

  • $\begingroup$ That would more fit as a comment. I would expect a more constructive answer, with at least a link or an example... $\endgroup$ – jpo38 May 2 '17 at 11:53
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    $\begingroup$ @jpo38 Meh, comments aren't really to give you information. Is there anything in particular that is unconstructive about the answer or doesn't address your concern? Without more information on what is giving you issues in the problem, I can't really go into more detail, especially because I don't think it's appropriate to actually work through a solution before you have shown any of your work (this is somewhat in line with the "homework-type question" policy on this site). $\endgroup$ – JMac May 2 '17 at 11:57
  • $\begingroup$ I understand, even if I did not flag that "homework", it was added by someone else. It's definitely not a homework, more something that came out in a friendly discussion and could not find an answer...I'll edit my post with what I have so far $\endgroup$ – jpo38 May 2 '17 at 12:20
  • $\begingroup$ @jpo38 It falls under homework-and-exercises due to the nature of it. That said, if you just explain what my answer doesn't tell you I would be more able to help. I feel I addressed what your question was really asking. $\endgroup$ – JMac May 2 '17 at 12:24
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    $\begingroup$ @jpo38 That looks correct. You could also probably find a tabulated value for water at 40°C instead of using that formula. It really is that simple though (easier if you use tabulated values instead of that formula). That is what I was trying to say. I also don't like giving direct examples for answers in case someone is just trying to get us to do their homework. Plus regardless of your level of physics, it's easier to understand when you can come to the conclusion yourself. Also note that these equations don't really say anything about how fast the water will gain heat energy. $\endgroup$ – JMac May 2 '17 at 12:38

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