heat to boil 1 gram water:
mcdeltaT+ Lv*m
1*1*(100-22)+540*1
heat to boil 1 gram Hg
mcdeltatT+Lv*m
1*.03(357-22)+65*1
do the math.
The specific heat of mercury is .03cal/g*C, and it's boiling point is 357*C. The specific heat of water is 1 cal/g*C. It takes 65 calories of energy to vaporize one gram of mercury and 540 calories to vaporize one gram of water. If both substances begin at about 22*C does it take more energy to boil a gram of mercury or a gram of water? If someone can give me the formula to use, I can try to figure it out myself! Thanks
2 answers
Energy required per gram =
Q = (heat of vaporization) + (Tboiling - Tinitial)*(specific heat)
Compute and compare that quantity for both liquids.
For mercury,
Q = 65 + 335*0.03
For water
Q = 540 + 78*1.0
Water's heat requirement is much higher
Q = (heat of vaporization) + (Tboiling - Tinitial)*(specific heat)
Compute and compare that quantity for both liquids.
For mercury,
Q = 65 + 335*0.03
For water
Q = 540 + 78*1.0
Water's heat requirement is much higher
1 answer