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Thermodynamics problem -- Find the mass of water that vaporizes

  • Thread starter Alice7979
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  • #1
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Homework Statement



Find the mass of water that vaporizes when 3.02 kg of mercury at 223 °C is added to 0.347 kg of water at 83.4 °C.

Homework Equations


Mercury Q = Water Q + Water mL
mc∆t= mc∆t + mL

The Attempt at a Solution


mass of vapor = ((3.02)(140)(223-tfinal) - (4186)(.347)(change in temperature))/100

I don't know the change in temperatures for sure, all the ones i have tried are wrong.
 

Answers and Replies

  • #2
Nathanael
Homework Helper
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all the ones i have tried are wrong.
It’s not a guessing game! The water will heat up to boiling temperature, then any additional energy will not raise the temperature (it instead goes into vaporization).

mc∆t= mc∆t + mL
This equation could solve the problem, if you have the correct subscripts (there are two c’s, two ∆t‘s, and three masses involved).
 
  • #3
19,962
4,109

Homework Statement



Find the mass of water that vaporizes when 3.02 kg of mercury at 223 °C is added to 0.347 kg of water at 83.4 °C.

Homework Equations


Mercury Q = Water Q + Water mL
mc∆t= mc∆t + mL

The Attempt at a Solution


mass of vapor = ((3.02)(140)(223-tfinal) - (4186)(.347)(change in temperature))/100

I don't know the change in temperatures for sure, all the ones i have tried are wrong.
Although not state explicitly, this assumes that the water is in contact with air at 1 atm, so the evaporation takes place at 100 C (the final temperature).
 
  • #4
36
2
It’s not a guessing game! The water will heat up to boiling temperature, then any additional energy will not raise the temperature (it instead goes into vaporization).


This equation could solve the problem, if you have the correct subscripts (there are two c’s, two ∆t‘s, and three masses involved).
mass of vapor = ((3.02)(140)(223-100) - (4186)(.347)(100-92.8))/100 = 540.6

That's what i tried but i still get it wrong, is it the change in temperature i have wrong?
 
  • #5
19,962
4,109
How much heat does the mercury give up in cooling from 223 to 100 C? How much of this heat must be used to heat the water up from 92.8 C to 100 C? The rest of the heat is used to vaporize some of the water.
 
  • #6
Nathanael
Homework Helper
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mass of vapor = ((3.02)(140)(223-100) - (4186)(.347)(100-92.8))/100 = 540.6

That's what i tried but i still get it wrong, is it the change in temperature i have wrong?
It would be nice, in the future, if you provided all relevant constants, so that I don’t have to look up everything.

It looks like you’re saying the energy of vaporization is 100 J/kg ?
Also I thought it started at 83.4 degrees not 92.8?
 
  • #7
36
2
It would be nice, in the future, if you provided all relevant constants, so that I don’t have to look up everything.

It looks like you’re saying the energy of vaporization is 100 J/kg ?
Also I thought it started at 83.4 degrees not 92.8?
Yes, I did have that wrong but I found it. Thanks
 

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