Answer:
the mass of steam at 100°C must be mixed is 150 g
Explanation:
given information:
ice's mass, [tex]m_{i}[/tex] = 490 g = 0.49 kg
steam temperature, T = 100°C
liquid water temperature, T = 89.0°C
specific heat of water, [tex]c_{w}[/tex] = 4186 J/kg.K = 4.186 kJ/kg.K
latent heat of fusion, [tex]L_{f}[/tex] = 333 kJ/kg
latent heat of vaporization, [tex]L_{v}[/tex] = 2256 kJ/kg
first, we calculate the heat of melted ice to water
Q₁ = [tex]m_{i} L_{f}[/tex]
where
Q = heat
[tex]m_{i}[/tex] = mass of the ice
[tex]L_{f}[/tex] = latent heat of fusion
thus,
Q₁ = [tex]m_{i} L_{f}[/tex]
= 0.49 x 333
= 163.17 kJ
then, the heat needed to increase the temperature of water to 89.0°C
Q₂ = [tex]m_{i}[/tex] [tex]c_{w}[/tex] (89 - 0), the temperature of ice is 0°C
[tex]c_{w}[/tex] = specific heat of water
so,
Q₂ = [tex]m_{i}[/tex] [tex]c_{w}[/tex] (89 - 0)
= 0.49 x 4.186 x 89
= 182.55 kJ
so, the heat absorbed by the ice is
Q = Q₁ + Q₂
= 163.17 + 182.55
= 345.72 kJ
the temperature of the steam is 100°C, so the mass of the steam is
Q = [tex]m_{s}[/tex][tex]L_{v}[/tex] + [tex]m_{s}[/tex][tex]c_{w}[/tex] (100 - 89)
Q = [tex]m_{s}[/tex]([tex]L_{v}[/tex] + [tex]c_{w}[/tex] (11))
[tex]m_{s}[/tex] = Q/ [[tex]L_{v}[/tex] + [tex]c_{w}[/tex] (11)]
= 345.72/ [2256 + (4.186 x 11)]
= 0.15 kg
= 150 g
