Benzene (C6H6) is an organic compound that boils at 80.1°C. If 100.0 grams of liquid benzene at 18.5°C is given 32.72 kJ of energy, what is the final temperature (in °C) of the benzene? The ΔHvaporization for benzene = 31.0 kJ/mol and the specific heat of liquid benzene = 2.18 J/gK.
I believe I did this right. But don’t take my word for it. So I believe you are supposed to convert kJ to Joules, as welll as J/gK to J/gC
0.03272 Joules=100.0g x -7.98J/gC x (T_final -18.5Celsius)
and then you divide both sides by 100.0g and -7.98 J/gC and you should get
-0.0409Celsius = T_final - 18.5 Celsius and then add 18.5 celsius to both sides and I got 18.5 Celsius. I think I did it correct .