Question

Find the mass of water that vaporizes when 3.28 kg of mercury at 227 °C is added to 0.233 kg of water at 93.6 °C.

Answer 1

Assuming that only part of water vaporizes, mixture of water and mercury will be at 100 C⁰.

Heat lost by mercury =3280 X .14 X (227-100) ; 0.14 J/gm specific heat of mercury

=58318.4 J

heat required to raise the tem. of water to 100 C⁰? =233 X 4.186 X (100-93.6) ; 4.186 J/gm specific heat of water

=6242.2 J

heat remaining to vaporize water is (58318.4-6242.2) J=52076.2 J

latent heat of vaporization 2230 J/gm

amount of water vaporized is =(52076.2/ 2230) gm

=23.4 gms

mass of water that vaporises is 23.4 gms