The problem involves calculating the mass of water that vaporizes when a specific mass of mercury at a high temperature is added to a smaller mass of water at a lower temperature. The context is thermodynamics, specifically focusing on heat transfer and phase changes.
Some participants have provided guidance on the heat transfer process and the need to account for the boiling point of water. There is an ongoing exploration of the correct values and assumptions, with no clear consensus reached on the solution.
There is mention of the need for relevant constants and clarification on initial temperatures, indicating that some information may be missing or unclear. Participants are also questioning the energy of vaporization used in calculations.
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?