Basic thermodynamic chemistry, heat transfer

AI Thread Summary
The discussion centers on a thermodynamics homework problem involving the heat transfer from steam to solid benzene. The user is unsure about the equations used to calculate heat gained by benzene and heat lost by steam, particularly regarding the signs and the final temperatures. There is confusion about whether to include the cooling of vapor or just its condensation, leading to negative mass results in calculations. Clarification is sought on whether the equation should be Qgained = Qlost and how to properly account for the heat transfer without mixing conventions. The importance of consistent sign conventions in thermodynamic calculations is emphasized to avoid nonsensical results.
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Homework Statement



How many grams of steam at 100 deg. Celcius would be required to raise the temperature of 35.8 g solid benzene from 5.5 deg Celcius to 45.0 deg Celcius?
Assume that heat is only transferred from the steam (not the liquid water) and that the steam/water and benzene are separated by a glass wall and do not mix.

Homework Equations


Boiling point of benzene 5.5 C
Δfus of benzen is 9.87 kJ/mol
Specific heat of benzene is 1.63 J/g.C and 4.18 for water
ΔHvap for steam at 100 C is 40.7 kJ/mol

im not sure about this part, since i can't find it in my book.

so the total heat of the system Qtotal =0= Qgained+Qlost
or Qgained = -Qlost

Qgained = molesbenzen ×ΔH+mbenzen×Cbenzen×(45-5.5)

Qlost = moleswater×ΔHvap + mwater×C×water×(45-100)

The Attempt at a Solution


So it seem like my equation above are wrong
because m comes out as a negative value, which made no sense.
I am pretty sure I got the heat gained correct.
I just uncertain about the heat lost of the steam.

Please give me some help.
Thanks for your time
 
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Wording is confusing. I guess you have to assume water vapor cooled down to 100 deg C and condensed, but the final temperature of water was 100 deg C.
 
it is the exact wording from the book, and this is just general chemistry, so I don't think there is any complexity in the problem.

so for the heat lost part, if the final temperature is 100 then the initial is 45?
and still, if it is (100-45) based on the above equation I cannot get m as a positive value?
 
Oops, sorry, there is no cooling of the vapor, only condensation.
 
So what should I do with this problem now?
Is my equation correct?
or should it be Qgained = Qlost instead?
 
Just leave the mcΔT part for water.
 
What about the negative sign in the equation Qgained = -Qlost
Should it be there?
even though we take out the mcΔT of water, with the negative sign, the answer still comes out to be negative =(
 
This sign thing can be confusing, as it can be done in many ways. What is important is to not mix conventions and stick to one. Sorry, no time to give more elaborate explanation now.

Good thing is usually when you mix conventions you get a nonsense result, which makes it easy to spot something is wrong. Negative mass is a sure sign something went wrong.
 

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