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