The mass of water that boils away is 0.01012 kg.
We know that the heat lost by mercury, Q equals heat gained by water, Q'
-Q = Q'
where Q = mc(T - T') where m = mass of mercury = 1.94 kg, c = specific heat capacity of mercury = 139.5 J/kg °C, T' = initial temperature of mercury = 200 °C and T' = final temperature of mercury = 100 °C (since this is the temperature at which the water will boil).
Also, Q' = Q" + Q"' where Q" = m'c'(T - T") where m = mass of water = 0.05 kg, c = specific heat capacity of water = 4200 J/kg °C, T" = initial temperature of water = 80 °C and T' = final temperature of water = 100 °C (since this is the temperature at which the water will boil).
Also, Q'" = m"L where m" = mass of evaporated water and L = latent heat of vaporization of water = 2.26 × 10⁶ J/kg
So, -Q = Q'
-Q = Q" + Q"'
-mc(T - T') = m'c'(T - T") + m"L
Making m" subject of the formula, we have
m" = -[mc(T - T') + m'c'(T - T")]/L
Substituting the values of the variables into the equation, we have
m" = -[mc(T - T') + m'c'(T - T")]/L
m" = -[1.94 kg × 139.5 J/kg °C(100 °C - 200 °C) + 0.05 kg × 4200 J/kg °C (100 °C - 80 °C)]/2.26 × 10⁶ J/kg
m" = -[1.94 kg × 139.5 J/kg °C(-100 °C) + 0.05 kg × 4200 J/kg °C (20 °C]/2.26 × 10⁶ J/kg
m" = -[270.63 J/ °C(-100 °C) + 210 J/ °C (20 °C)]/2.26 × 10⁶ J/kg
m" = -[-27063 J + 4200 J]/2.26 × 10⁶ J/kg
m" = -[-22863 J]/2.26 × 10⁶ J/kg
m" = 22863 J/2.26 × 10⁶ J/kg
m" = 10116.372 × 10⁻⁶ kg
m" = 0.010116372 kg
m" ≅ 0.01012 kg
So, the mass of water that boils away is 0.01012 kg
Learn more about mass of water that boils away here:
https://brainly.com/question/13188500
