2 kg of water is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg K and its densities are 999.9 kg/ms and 1000 kg/ms respectively at these temperatures. The atmospheric pressure is 10⁵ pascal. The change in the internal energy is

Heat supplied = m s $△$T   
              $=2×4200×4=33600\mathrm{J}$ = 33.6 KJ
Change in volume,
$△V=\frac{2}{999⋠9}−\frac{2}{1000}=0⋠2×{10}^{−6}$ $m^3$
Work done $= -P△V= -{10}^5×\left(0⋠2×{10}^{−6}\right)\mathrm{J}= -0⋠02 J$ J
Net increase in internal energy = Q - W
               $\begin{array}{l}=(33600+0⋠02)\mathrm{J}\\ =33⋠6\mathrm{kJ}+2×{10}^{−5}\mathrm{kJ}=\left(33⋠6+2×{10}^{−5}\right)\mathrm{kJ}\end{array}$