A wet cloth has been hung out to dry....

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Wet clothing at 0°C can dry through evaporation and freezing when exposed to dry wind, but calculating the fraction of water that remains as ice is challenging due to the difficulty in measuring the mass of air and the heat it supplies. The specific latent heats of fusion and vaporization are crucial for solving the problem, but the source of heat for evaporation at 0°C is questioned. It is noted that while the air temperature is the same, the dry wind aids in removing water vapor, which influences the drying process. The discussion highlights the need to understand the heat dynamics involved in both evaporation and freezing. Ultimately, the problem requires a deeper analysis of heat transfer mechanisms in this scenario.
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Homework Statement


Wet clothing at 0°C is hung out to dry when the air temperature is 0°C and there is a dry wind blowing. After some time, it is found that some of the water has evaporated and the remainder has frozen. Estimate the fraction of the water originally present in the clothing which remains as ice. (At 0°C, specific latent heat of fusion of ice = 336 kJkg-1, specific latent heat of vaporization of water = 2500 kJkg-1, specific heat of water = 4200 Jkg-1K-1)

Homework Equations


Q=msΔθ where Q=amount of heat,m=mass,s=specific heat ,Δθ=change of temperature.

The Attempt at a Solution


The problem can be solved by using the given formula.But I am facing problem that it is impossible to measure the mass of air and as a result, I also failed to compute the heat supplied by the dry air to the water.So how can I get rid of here?
 
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Akash47 said:

Homework Statement


Wet clothing at 0°C is hung out to dry when the air temperature is 0°C and there is a dry wind blowing. After some time, it is found that some of the water has evaporated and the remainder has frozen. Estimate the fraction of the water originally present in the clothing which remains as ice. (At 0°C, specific latent heat of fusion of ice = 336 kJkg-1, specific latent heat of vaporization of water = 2500 kJkg-1, specific heat of water = 4200 Jkg-1K-1)

Homework Equations


Q=msΔθ where Q=amount of heat,m=mass,s=specific heat ,Δθ=change of temperature.

The Attempt at a Solution


The problem can be solved by using the given formula.But I am facing problem that it is impossible to measure the mass of air and as a result, I also failed to compute the heat supplied by the dry air to the water.So how can I get rid of here?
Does air of 0°C supply heat to water also at 0°C? Where does the heat come that evaporates the water?
 
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ehild said:
Does air of 0°C supply heat to water also at 0°C? Where does the heat come that evaporates the water?
I am not sure whether the air of 0°C supply heat.But of course the heat comes from the dry air.But that doesn't solve my problem which I mentioned in my attempt of the solution.
 
Akash47 said:
I am not sure whether the air of 0°C supply heat.But of course the heat comes from the dry air.But that doesn't solve my problem which I mentioned in my attempt of the solution.
The dry wind removes the water vapor, and keeps the cloth at 0°C. But where does the heat come that evaporates the water? Think: what happens if water freezes?
 
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