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Thermodynamics

Question
CBSEENPH11026330

1 kg of water is converted into steam at the same temperature and at 1 atm (100 kPa). The density of water and steam are 1000 kgm-3 and 0.6 kgm-3 respectively. The latent heat of vaporisation of water is 2.25 x 106 J kg-1, What will be increase in energy?

  • 3 × 105 J

  • 4 × 106 J

  • 2.08 × 106 J

  • None of these

Solution

C.

2.08 × 106 J

Given:-

Latent heat of vaporization = 2.25 × 106 J kg-1

P = 1atm = 100kPa = 1 × 105 Pa 

Mass of the water = 1 kg

The volume of 1 kg water

⇒          V = 11000 m3

and volume of 1 kg steam 

⇒         V' = 10.6 m3

The increase in volume 

⇒           ΔV = ( V' - V )

⇒          ΔV  = 10.6 m3-  11000 m3

                   = ( 1.7 - 0.001 ) m3

⇒           ΔV = 1.699

⇒           ΔV ≈ 1.7 m3

The work done by steam is

      ΔW = P ΔV

      ΔW = ( 100 kPa ) (1.7 m3 )

      ΔW = 1.7 × 105 J

The change in internal energy

      ΔU = ΔQ - ΔW

           = 2.25 × 106 J - 1.7 × 105 J

      ΔU = 2.08 × 106 J