Explanation:
The given data is as follows.
Volume = 65.5 [tex]cm^{3}[/tex]
Temperature = [tex]100^{o}C[/tex]
Pressure = 760 torr or 1 atm
As we know that q = [tex]mL_{f}[/tex]
where, m = mass
[tex]L_{f}[/tex] = latent heat of fusion
As density of water is 1.0 g/ml. Hence, mass of water in liquid state is calculated as follows.
mass = density × volume
= [tex]1.00 g/ml \times 65.5 ml[/tex] (as 1 ml = 1 [tex]cm^{3}[/tex])
= 65.5 g
As heat of fusion of ice is 333.55 J/g. Hence, calculate the heat energy as follows.
q = [tex]mL_{f}[/tex]
= [tex]65.5 g \times 333.55 J/g[/tex]
= 21847.52 J
And, heat of vaporization of water is 2257 J/g.
Also, q = [tex]mL_{v}[/tex]
where, [tex]L_{v}[/tex] = heat of vaporization
Therefore, we will calculate the mass of steam as follows.
q = [tex]mL_{v}[/tex]
21847.52 J = [tex]m \times 2257 J/g[/tex]
m = 9.68 g
It is known that density of steam water is 0.0006 g/ml. Hence, calculate the volume of steam as follows.
Volume = [tex]\frac{mass}{Density}[/tex]
= [tex]\frac{9.68 g}{0.0006 g/ml}[/tex]
= 16133.33 ml
or, = 16.13 L (as 1 L = 1000 ml)
Thus, we can conclude that 9.68 g of steam condenses will release the same amount of energy as 65.5 g of freezing liquid water and its volume is 16.13 L.
