Find the mass of water that vaporizes when 4.95 kg of mercury at 210 °C is added to 0.162 kg of water at 92.0 °C. Solution let (m) mass of water evaporates heat lost by mercury = heat gained by water (sensible heating) + heat needed for phase change ------------ 4.95*139.5[210 - 100] = (0.162) *4182*(100-92)+ m*(2.23*10^6) 75957.75 = 5419.872 + m*2.23*10^6 m = 0.03163 kg (= 31.63g) [The answer may vary a little depending on the values of specific heats used. Use the values given to you] .

1. Find the mass of water that vaporizes when 4.95 kg of mercury at 210 °C is added to 0.162 kg
of water at 92.0 °C.
Solution
let (m) mass of water evaporates
heat lost by mercury = heat gained by water (sensible heating) + heat needed for phase change
------------
4.95*139.5[210 - 100] = (0.162) *4182*(100-92)+ m*(2.23*10^6)
75957.75 = 5419.872 + m*2.23*10^6
m = 0.03163 kg (= 31.63g)
[The answer may vary a little depending on the values of specific heats used. Use the values
given to you]