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