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
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 =
and volume of 1 kg steam
⇒ V' =
The increase in volume
⇒ ΔV = ( V' - V )
⇒ ΔV = m3 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