Calculate the increase in the internal energy of 10g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg m–3. The specific heat capacity of water = 4200 J kg–1 °C–1 and the latent heat of vaporization of water = 2.25 × 106 J kg–1.
Given
The density of steam ρ’= 0.6 kg m–3
Mass of water m=10g =0.010kg
Specific heat capacity of water c = 4200 J kg–1 °C–1
latent heat of vaporization of water L = 2.25 × 106 J kg–1.
Pressure P =100kPa =100×105Pa
Change in temperature ΔT= (100-0) oC =100oC
Density of water ρ =1000 kg m-3
We know that specific heat capacity is given by
Where ΔQ = heat supplied to the system
Therefore, ΔQ= cmΔT
Also, ΔQ =mL
Where m= mass of the substance
L=latent heat
Therefore, ΔQ=mL + cmΔT
=0.010×2.25 × 106 + 4200×0.01×100
=22500+4200
=26700J
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
Also,
ΔW=105×0.01699=1699 J
From first law of thermodynamics, we know that,
ΔQ=ΔU+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔU=ΔQ-ΔW
=26700-1699
=25001J
Thus, change in internal energy is 25001J.