Calculate work needed to evaporate water

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Discussion Overview

The discussion revolves around calculating the work needed to evaporate water from washed clothes in a well-isolated room at a temperature of 75°F. Participants explore the energy requirements for evaporation, considering factors such as latent heat and temperature effects.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant suggests using the formula Q = m h_{fg} to calculate the energy required for evaporation, where Q is the energy, m is the mass of water, and h_{fg} is the latent heat of vaporization.
  • Another participant emphasizes that the initial temperature of the water (75°F) must be considered, arguing that it affects the energy required for evaporation compared to water at different temperatures.
  • A further contribution proposes including sensible heat in the calculation with the formula q = m h_{fg} + m c_{p} ΔT, where c_{p} is the specific heat and ΔT is the temperature change from 75°F to 212°F.
  • One participant expresses a practical application of the calculation, aiming to determine the cost savings of line drying clothes versus using an electric dryer while maintaining a constant temperature in the house.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the initial temperature significantly alters the energy calculation for evaporation, as some argue for its inclusion while others focus solely on latent heat.

Contextual Notes

There are unresolved assumptions regarding the specific heat values and the precise conditions under which the calculations are made, as well as the impact of maintaining a constant room temperature on energy consumption.

pikiche
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how would you calculate the work needed to evaporate water from washed clothes, if the clothes r in a well isolated room with temperature of 75F
help pleasezzz
 
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The energy required is the mass of water multiplied by the laten heat of vaporization:

Q = m [itex]h_{fg}[/itex]

Q = energy required
m = mass of water
[itex]h_{fg}[/itex] = latent heat of vaporization

If you want to estimate how long it will take to evaporate naturally, search for "swimming pool evaporation equation."
 
edgepflow said:
The energy required is the mass of water multiplied by the laten heat of vaporization:

Q = m [itex]h_{fg}[/itex]

Q = energy required
m = mass of water
[itex]h_{fg}[/itex] = latent heat of vaporization

If you want to estimate how long it will take to evaporate naturally, search for "swimming pool evaporation equation."

No, he said it starts off at 75 degrees. Surely that has to be taken into account. I mean, it CAN'T take the same amount of energy to evaporate water at 1 degree above freezing as to do it to water at 1 degree below boiling.
 
phinds said:
No, he said it starts off at 75 degrees. Surely that has to be taken into account. I mean, it CAN'T take the same amount of energy to evaporate water at 1 degree above freezing as to do it to water at 1 degree below boiling.
The sensible heat can be included as follows:

q = m [itex]h_{fg}[/itex] + m [itex]c_{p}[/itex][itex]\Delta[/itex]T

cp = specific heat
[itex]\Delta[/itex]T = temperature change = 212 F - 75F
 
thank you for the reply guys, see what I am trying to do is actually calculate how much money would u save from line drying clothes vs using an electric dryer. knowing that u have an ac/ heater at the houseto keep the temperature at 75. so i was thinking that after calculating the heat to evaporate the water from the clothes wouldn't be the same amount of energy that the ac/heater has to apply to maintain the temperature at 75?
 

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