A cube of side $$5\,cm$$  made of iron and having a mass of $$1500\,g$$   is heated from $${25^ \circ }C$$  to $${400^ \circ }C.$$  The specific heat for iron is $$0.12\,cal/{g^ \circ }C$$   and the coefficient of volume expansion is $$3.5 \times {10^{ - 5}}{/^ \circ }C,$$    the change in the internal energy of the cube is (atm pressure $$ = 1 \times {10^5}\,N/{m^2}$$   )

Solution :
$$\eqalign{ & Q = mC\Delta T = 1.5 \times 0.12 \times 4200 \times \left( {400 - 25} \right) \cr & = 2.83 \times {10^5}\,J \cr & W = P\left( {\Delta V} \right) = P\left( {V\gamma \Delta T} \right) \cr & = {10^5} \times {\left( {5 \times {{10}^{ - 2}}} \right)^3} \times 3.5 \times {10^{ - 5}} \times 375 = 0.164\,J \cr & {\text{Thus}}\,\,Q = \Delta U + W \cr & {\text{or}}\,\,2.83 \times {10^5} = \Delta U + 0.164;\Delta U = 282\,kJ \cr} $$