Q. 205.0( 1 Vote )

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.

Answer :

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.


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