Arthur Clarke wrote an interesting short story called "A Slight Case of Sunstroke." Disgruntled football fans came to the stadium one day equipped with mirrors and were ready to barbecue the referee if he favored one team over the other. Imagine the referee to be a cylinder filled with water of mass \(60.0 \mathrm{~kg}\) at \(35.0^{\circ} \mathrm{C}\). Also imagine that this cylinder absorbs all the light reflected on it from 50,000 mirrors. If the heat capacity of water is \(4.20 \cdot 10^{3} \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right),\) how long will it take to raise the temperature of the water to \(100 .{ }^{\circ} \mathrm{C}\) ? Assume that the Sun gives out \(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2},\) the dimensions of each mirror are \(25.0 \mathrm{~cm}\) by \(25.0 \mathrm{~cm},\) and the mirrors are held at an angle of \(45.0^{\circ}\)

First, we need to calculate the total energy needed to raise the temperature of the water to \(100^{\circ} \mathrm{C}\). We can use the formula for the heat capacity to find this out: \(Q = mc\Delta T\) Where: \(Q\) = Total energy needed (Joules) \(m\) = Mass of the water (\(60.0 \mathrm{~kg}\)) \(c\) = Heat capacity of water (\(4.20 \cdot 10^{3} \mathrm{~J} / \left(\mathrm{kg}^{\circ} \mathrm{C}\right)\)) \(\Delta T\) = Change in temperature (\(100^{\circ} \mathrm{C} - 35.0^{\circ} \mathrm{C}\)) \(Q = 60.0 \mathrm{~kg} \times 4.20 \cdot 10^{3} \mathrm{~J} / \left(\mathrm{kg}^{\circ} \mathrm{C}\right) \times (100^{\circ} \mathrm{C} - 35.0^{\circ} \mathrm{C})\) \(Q = 60.0 \mathrm{~kg} \times 4.20 \cdot 10^{3} \mathrm{~J} / \left(\mathrm{kg}^{\circ} \mathrm{C}\right) \times (65^{\circ} \mathrm{C})\) \(Q = 16.38 \cdot 10^{6} \mathrm{~J}\) So, the total energy needed to raise the temperature of the water to \(100^{\circ} \mathrm{C}\) is \(16.38 \cdot 10^{6} \mathrm{~J}\).#End_tag#

Next, we need to find the total power received by the cylinder per second from all the mirrors. We are given the dimensions of each mirror and the angle at which they are held, as well as the power output from the sun. First, calculate the area of each mirror: \(A = lw\) Where: \(A\) = Area (\(\mathrm{m}^{2}\)) \(l\) = Length (\(0.25 \mathrm{~m}\)) \(w\) = Width (\(0.25 \mathrm{~m}\)) \(A = 0.25 \mathrm{~m} \times 0.25 \mathrm{~m} = 0.0625 \mathrm{~m}^{2}\) Then, since we have 50,000 mirrors, the total area of all mirrors collectively will be: \(A_{total} = n \times A\) Where: \(n\) = Number of mirrors (50,000) \(A_{total} = 50{,}000 \times 0.0625 \mathrm{~m}^{2} = 3125 \mathrm{~m}^{2}\) Now, we can find the total power received by the cylinder per second: \(P_{total} = I \times A_{total} \times \cos{\theta}\) Where: \(I\) = Intensity of sunlight (\(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2}\)) \(\theta\) = Angle of the mirrors (\(45^{\circ}\)) \(P_{total} = 1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2} \times 3125 \mathrm{~m}^{2} \times \cos{(45^{\circ})}\) \(P_{total} = 1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2} \times 3125 \mathrm{~m}^{2} \times 0.707\) \(P_{total} = 2.21 \cdot 10^{6} \mathrm{~W}\) So the total power received per second by the cylinder from the mirrors is \(2.21 \cdot 10^{6} \mathrm{~W}\).#End_tag#

Finally, we can find the time needed to raise the temperature of the water to \(100^{\circ} \mathrm{C}\) using the total energy needed and the total power received per second: \(t= \frac{Q}{P_{total}}\) Where: \(t\) = Time needed (seconds) \(t= \frac{16.38 \cdot 10^{6} \mathrm{~J}}{2.21 \cdot 10^{6} \mathrm{~W}}\) \(t \approx 7.42 \mathrm{~seconds}\) Therefore, it will take approximately 7.42 seconds to raise the temperature of the water in the cylinder to \(100^{\circ} \mathrm{C}\).