Calculate work needed to evaporate water

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SUMMARY

The work needed to evaporate water from washed clothes in a well-isolated room at 75°F can be calculated using the formula Q = m h_{fg}, where Q is the energy required, m is the mass of water, and h_{fg} is the latent heat of vaporization. Additionally, to account for the initial temperature of the water, the sensible heat must be included in the calculation: 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. This comprehensive approach allows for accurate estimations of energy requirements for evaporation and can be used to compare costs between line drying and using an electric dryer.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically latent heat of vaporization
  • Familiarity with the specific heat capacity of water
  • Basic knowledge of energy calculations in physics
  • Ability to perform unit conversions between Fahrenheit and Celsius if necessary
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  • Research the "swimming pool evaporation equation" for natural evaporation rates
  • Learn about the specific heat capacity of water and its implications in energy calculations
  • Investigate the cost analysis of line drying versus electric dryer usage
  • Explore thermodynamic principles related to energy conservation in closed systems
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Students studying thermodynamics, homeowners considering energy-efficient laundry options, and anyone interested in the physics of evaporation and energy consumption.

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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 h_{fg}

Q = energy required
m = mass of water
h_{fg} = 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 h_{fg}

Q = energy required
m = mass of water
h_{fg} = 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 h_{fg} + m c_{p}\DeltaT

cp = specific heat
\DeltaT = 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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