It is desired to harden a cylindrical steel rod by changing its structural form from austenite to martensite. In order to do this, the steel must be cooled rapidly by a process called quenching. The hot steel, which is at a temperature of \(900^{\circ} \mathrm{C}\), must be cooled to below its eutectoid point, which is at \(650^{\circ} \mathrm{C}\).

a. Assuming the approximation for lumped capacitance is valid, calculate the time taken to cool the steel rod from \(900^{\circ} \mathrm{C}\) to \(650^{\circ} \mathrm{C}\). The length of the rod is \(10 \mathrm{~cm}\), and the diameter is \(4 \mathrm{~cm}\). The density of the rod is \(8000 \mathrm{~kg} / \mathrm{m}^{3}\) and the specific heat capacity is \(480 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\). The heat-transfer coefficient between the cylinder and water is \(300 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

b. If the volume of the tank is \(0.01 \mathrm{~m}^{3}\), find the approximate change in the temperature of the water. The initial temperature of the water is \(30^{\circ} \mathrm{C}\).