The problem involves calculating 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. The relevant equations include the heat transfer equation mc∆t and the relationship between the heat lost by mercury and the heat gained by water. The final calculation for the mass of vapor resulted in 540.6 grams, but the user struggled with determining the correct change in temperature and constants involved. The discussion emphasizes the importance of accurate temperature values and constants for solving thermodynamic problems.
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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).Alice7979 said:all the ones i have tried are wrong.
This equation could solve the problem, if you have the correct subscripts (there are two c’s, two ∆t‘s, and three masses involved).Alice7979 said:mc∆t= mc∆t + mL
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).Alice7979 said: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.
Nathanael said: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).
It would be nice, in the future, if you provided all relevant constants, so that I don’t have to look up everything.Alice7979 said: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?
Yes, I did have that wrong but I found it. ThanksNathanael said: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?