Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J kg–1 K–1 and its densities at 0°C and 4°C are 999.9 kg m–3 and 1000 kg m–3 respectively. Atmospheric pressure = 105 Pa.
Given
Mass of water m = 2kg
Change in temperature 
T =4℃ -0℃ =4℃
Specific heat capacity of water c =4200 J kg–1 K–1
Density of water at 0°C=999.9 kg m–3
Density of water at 4°C=1000 kg m–3
Atmospheric pressure P = 105 Pa.
We know that specific heat capacity is given by
Where ΔQ = heat supplied to the system
Therefore, ΔQ=cmΔT
=4200×2×4=33600J
We know that work done by the gas is given as
ΔW=PΔV
Where ΔV =change in volume
P =pressure
Also,
Volume at 0oC V1
Similarly, volume at 4oC V2
∴ ΔW=P(V2-V1)
From first law of thermodynamics, we know that,
Q=
U+ΔW
Where ΔQ=heat supplied to the system
ΔU=change in internal energy
ΔW=work done by the system
ΔU=ΔQ-ΔW
=33600-(-0.02)
=33599.98J
Thus, change in internal energy is 33599.98J.