During a reversible isobaric non-flow process with \(p=1.5\) bar the properties of the system change from \(0.25 \mathrm{~m}^{3} / \mathrm{kg}, 10^{\circ} \mathrm{C}\) to \(0.45 \mathrm{~m}^{3} / \mathrm{kg}, 240^{\circ} \mathrm{C}\). The specific heat of the fluid is given by \(c_{p}=[1.6+80 /(t+40)] \mathrm{kJ} / \mathrm{kg} .{ }^{\circ} \mathrm{C}\). Calculate \((a)\) heat added \(/ \mathrm{kg},(b)\) work done \(/ \mathrm{kg},(c)\) change in internal energy \(/ \mathrm{kg}\), and (d)change in enthalpy \(/ \mathrm{kg}\).