Question

What volume and mass of steam at 100 Celsius and 760. torr (or 1 atm) would release the same amount of energy as heat during condensation as 65.5cm^3 of liquid water would release during freezing?

Answer 1

Explanation:

The given data is as follows.

          Volume = 65.5 $cm^{3}$

          Temperature = $100^{o}C$

          Pressure = 760 torr or 1 atm

As we know that q = $mL_{f}$  

where,    m = mass

            $L_{f}$ = 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

                            = $1.00 g/ml \times 65.5 ml$            (as 1 ml = 1 $cm^{3}$)

                           = 65.5 g

As heat of fusion of ice is 333.55 J/g. Hence, calculate the heat energy as follows.

                    q = $mL_{f}$  

                       = $65.5 g \times 333.55 J/g$

                       = 21847.52 J

And, heat of vaporization of water is 2257 J/g.

Also,                q = $mL_{v}$

where,      $L_{v}$ = heat of vaporization

Therefore, we will calculate the mass of steam as follows.

                        q = $mL_{v}$

                     21847.52 J = $m \times 2257 J/g$            

                       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 = $\frac{mass}{Density}$

                              = $\frac{9.68 g}{0.0006 g/ml}$

                              = 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.

Answer 2

  
As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O 
now

Heat of fusion of water = 79.8 cal/g 
and

Heat of vaporization of water = 540 cal/g 

Atomic weight of water : H=1 O=16 H2O=18 
now by calculating and putting values

65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam) 

9.68gH2O x 1molH2O/18gH2O x 22.4LH2O/1molH2O = 12.0 L H2O
hope it helps