Posted by **Angel** on Wednesday, January 25, 2012 at 8:31pm.

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

- physics -
**bobpursley**, Wednesday, January 25, 2012 at 8:36pm
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.

- physics -
**drwls**, Wednesday, January 25, 2012 at 8:41pm
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

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