Answer to Question #84488 in Molecular Physics | Thermodynamics for Jake

Question

Find the mass of water that vaporizes when 2.1kg of mercury at 205 C is added to 0.11kg of water at 80 C.

Answer 1

You need heat energy to first heat the water from 80oC to its boiling point of 100oC, and then additional heat energy to vaporize the water into steam. The source of that energy is the hot, 205oC Mercury which is thrown into the 80oC water, so the Mercury will be cooled from 205oC down to 100oC. If it has insufficient heat energy to give up, then not all the water can be vaporized.

The Mercury can supply Qm = c1M1ΔT1 = (140) (2.1) (205 - 100) = 30.87 KJ of heat energy

The water needs Qw = C2 M2 ΔT2 = (4.184 K) (0.11 Kg) (100 - 80) = 9.20 KJ of heat energy to raise all its mass to 100oC

Thus, 30.87 KJ - 9.20 KJ = 21.67 KJ is available for water vaporization

Mw = 21.67 KJ/L, where L is the heat of vaporization for water = 2260 KJ/Kg

So Mw = 21.67/2260 = 0.00959 Kg of water can be vaporized.