Heat & Phase change

Question

Find the mass of water that vaporizes when 2.1 kg of mercury at 205C is added to .11kg of water at 80C.

 

The answer in the back of the book is .00949 kg, which I cannot figure out how they get.

 

thank you

john

Answer 1

John,

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

 

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

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

 

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

 

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

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

 

Again, this answer is slightly off from the number you quote, probably a small difference in the constants used or rounding in intermediate calculations.

 

Note, available energy is first used to raise all the water to 100^oC or some lower temperature.  If you had insufficient heat energy (say the mass of mercury was too low), then the water temperature would have settled below 100^oC with no vaporization. You can't have part of the water vaporized and the rest colder than 100^oC.