Answer:
Condensed steam, M = 0.091 g
The remaining steam = 199.9 g
Given:
mass of lead, m = 50 g
temperature, T = -10[tex]{^\circ}[/tex]
mass of steam, m' = 200 g
temperature, T' = 105[tex]{^\circ}[/tex]
Solution:
Now,
latent heat of vaporization of steam, L = 2260 J/g
specific heat of lead, [tex]C_{l}[/tex] = 0.128 J/g-K
specific heat of steam, [tex]C_{s}[/tex] = 1.996 J/g-K
Now, absorbed heat of lead to reach [tex]0^{\circ][/tex]:
q = m[tex]C_{l}\Delta T[/tex]
q = [tex]50\times 0.128\times (100-(-10))[/tex]
q = [tex]50\times 0.128\times 110[/tex]
q = 704 J
Now, to reach 100[tex]{^\circ}[/tex] the amount of heat released by steam:
q' = m'[tex]C_{s}\Delta T[/tex]
q' = [tex]50\times 1.996\times (105 - 100)[/tex]
q' = 499 J
Now,
Let the condensed part of the steam have mass M, then:
ML = q - q'
M = [tex]\frac{q - q'}{L}[/tex]
M = [tex]\frac{704 - 499}{2260}[/tex]
M = 0.091 g
The remaining steam = 200 - 0.091 = 199.9 g
