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.

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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