Heat required to evaporate water.

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To calculate the heat required to evaporate 1 kg of water at temperatures below 100 degrees Celsius, the latent heat of vaporization must be considered, which varies with temperature. At a constant temperature of 50 or 70 degrees Celsius, the heat required will not be equal to the latent heat at 100 degrees Celsius. The heat transfer equation Q = mcΔt is typically used for raising temperature, but for evaporation, the latent heat specific to the temperature must be used. Without access to steam tables or psychrometric charts, one can refer to online resources for latent heat values at various temperatures. Understanding these principles is essential for accurate calculations in thermal processes.
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If 1 kg water is required to be evaporated, not at 100 degree Celsius, how can I calculate the heat required.

Let say a constant temperature of 50 degree Celsius is maintained, then what will be the scenario, and what if the constant temperature of 70 degree Celsius is maintained.
 
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Usually, when calculating the heat required, use the heat transfer equation:

Q= mcΔt

Q=energy, or heat
m=mass
c=specific heat capacity
delta t= change in time
 
Last edited:
makovx said:
Usually, when calculating the heat required, use the heat transfer equation:

Q= mcΔt

Q=energy, or heat
m=mass
c=specific heat capacity
delta t= change in time

Delta T, in my understanding is the change in temperature.

And this equation is for calculating heat required to raise the temperature by certain degree Celsius of the given material.

I want to keep the temperature constant, that is, supplying the heat through a constant temperature air.

I wonder that will the heat required will be equal to the latent heat of vaporization of water (as it is on 100 degree Celsius), or should there be any difference.
 
You do not have access to steam tales or a psychrometric chart or a Mollier diagram to determine the enthalpy change from liquid water to vapour at a set temperature?
 
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