How much heat is being drawn when water vaporize?

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Discussion Overview

The discussion revolves around the energy involved in the vaporization of water, specifically how to calculate the heat drawn from a bowl of water when 10 grams evaporate. Participants explore concepts related to latent heat, the heat of vaporization, and the effects of air temperature and humidity on this process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about calculating the energy lost from water during evaporation, despite understanding latent heat and heat of vaporization.
  • Another participant suggests that the heat of vaporization can be calculated by multiplying the mass of evaporated water by the specific heat of vaporization.
  • Some participants note that the heat exchange involves both the water losing energy and the air gaining energy, complicating the calculation of energy solely lost from the water.
  • There is a discussion about the role of air temperature and humidity in the cooling process of the water, with mentions of how the rate of evaporation is influenced by these factors.
  • One participant questions how to account for the energy carried by water molecules and whether to consider absolute energy or changes in energy for calculations.
  • Another participant emphasizes that the energy for vaporization ultimately comes from the air, and the latent heat of vaporization remains constant regardless of the source of energy.
  • There are attempts to apply formulas for heat transfer, but some participants express uncertainty about the correct application and the variables involved.
  • One participant raises a concern about the realism of the thought experiment, suggesting that the water cannot evaporate without an external heat source.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement regarding the calculations and concepts involved in vaporization. While some agree on the basic principles of latent heat, others challenge the assumptions and methods used in calculations, indicating that the discussion remains unresolved.

Contextual Notes

Participants highlight the complexity of the problem, including the need to consider the effects of air temperature, humidity, and the source of energy for vaporization. There are unresolved questions about the correct application of formulas and the assumptions made in the calculations.

Who May Find This Useful

This discussion may be useful for individuals interested in thermodynamics, heat transfer, and the principles of phase changes in water, particularly in the context of evaporation and energy calculations.

  • #31
enth,

When you say a bowl of water is at such-and-such a temperature, that effectively specifies the average kinetic energy in each molecule. At any instant, though, some will have more, some less, according to a standard probability distribution.
The more energetic ones may have enough energy to escape the bowl and become vapour.
At the same time, the water vapour molecules in the air are bouncing around, gaining and losing energy. Some lose so much that they get caught by the water.

When these two processes are in balance, no net evaporation or condensation occurs.
Make the air a bit drier and fewer are available to recondense, so net evaporation does occur. Make the water a bit warmer and more are available to evaporate.
So at a given water temperature, there's a certain density of vapour above it (the saturation vapour pressure at that temperature) at which the system is in equilibrium.

In all cases, evaporation cools the water and condensation heats it.

The fan in your model is there purely to ensure a steady supply of air at the given temperature and humidity (otherwise the air just above the water gets saturated and cold). Similarly, the surface layer of the water will cool first, inhibiting further evaporation, but as long as it's above 4C convection should kick in and redistribute the heat.

Based on the above, you can do the calculation exactly as SophieC said.
temperature drop in bowl = heat required to vaporise lost water / thermal capacity of remaining water.

Once the water starts to cool, it will be cooler than the air above. This means some heat will flow from the air to the water. As a result, the temperature drop will not be quite as great.
 

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