Question

1.Find the mass of water that vaporizes when 2.03 kg of mercury at 222

Answer 1

1)

Mass of the water, Mw = 0.112 kg

Mass of the mercury , Mm = 2.03 kg

Initial temperature of the water, t1 = 72.8 deg c

Initial temeperature of mercury, t2 = 222 deg c

Specific heat of water, Cw= 4186 J/kg/K

Latent heat of water, L = 2.26 x 10^6 J/kg

Specific heat of mercury, Cm = 139 J/kg/K

Let

Resultant temperature of the mixture = 100 deg c

Solution:

From the principle of method of mixtures,

Heat gained by water = Heat lost by mercury

Mw Cw ( t - t1 ) + x * L = Mm Cm ( t2 - t )

0.112 * 4186 * 27.2 + x * 2.26 x 10^6 = 2.03 * 139 * 122

x * 2.26 x 10^6 = 2.1672 x 10^4

                        x = 0.00959 kg

                           = 9.59 g

2)

Q = m(2020)(2.4) + (m)(2.26 X 10^6) +(m)(4186)(100) + (m)(3.33 X 10^5)

Q = 3016448 m

energy = .5mv^2

(3016448)m = .5mv^2   (mass cancels)

v = 2456.19 m/s