Answer
See explanation
Work Step by Step
(a) To find $\Delta V = \beta \Delta TV_o $
$\Delta V = \beta \Delta TV_o $
$\Delta V = 5.1 \times 10^{-5} (^{o}C)^{-1}(90^oC - 20^oC)(2.00 \times 10^{-2} m)^3$
$\Delta V = 5.1 \times 10^{-5} (^{o}C)^{-1}(70^oC)(8.00 \times 10^{-6} m)^3$
$\Delta V = 2.86 \times 10^{-8} m^3$
(b) $W = p \Delta V $
$W = (1.01 \times 10^5 Pa) 2.86 \times 10^{-8} m^3 $
$W = 2.89 \times 10^{-3} J$
(c) $Q = mC_p \Delta T $
$Q = \rho V_oC_p\Delta T $
$Q = (8.90×10^3 kg/m^3)(8.00 \times 10^{-6} m)^3) (390 J/kg.K) (70 K)$
$Q = 1944 J$
(d) Change in internal energy, $\Delta U $ of the cube is the same as the heat added to the cube. Thi
$ΔU = Q - W$
$ΔU = 1944 J - 2.89 \times 10^{-3} J$
$ΔU = 1943.997 J $
(e) Under these conditions, the difference is not substantial, since W is much less than Q. This is observable that solid did much less work compared to condition in gas.
