Question
How many grams of methanol could be vaporized by the addition of 22.45kJ of heat?
Answers
Devron
Use the formula
q = m·ΔHv
where
q=heat energy=22.45kJ= 22.45 x 10^3J
m=mass=?
ΔHv=heat of vaporization=1,006 J/g
Solve for m,
q/ΔHv=m
(22.45 x 10^3J/1,006 J/g)=m
q = m·ΔHv
where
q=heat energy=22.45kJ= 22.45 x 10^3J
m=mass=?
ΔHv=heat of vaporization=1,006 J/g
Solve for m,
q/ΔHv=m
(22.45 x 10^3J/1,006 J/g)=m