[qpham26]

1. The problem statement, all variables and given/known data

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.


2. Relevant equations
Boiling point of benzene 5.5 C
Δ_fus of benzen is 9.87 kJ/mol
Specific heat of benzene is 1.63 J/g.C and 4.18 for water
ΔH_vap for steam at 100 C is 40.7 kJ/mol

im not sure about this part, since i cant find it in my book.

so the total heat of the system Q_total =0= Q_gained+Q_lost
or Q_gained = -Q_lost

with

Q_gained = moles_benzen ×ΔH+m_benzen×C_benzen×(45-5.5)

Q_lost = moles_water×ΔH_vap + m_water×C×_water×(45-100)

3. 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