A town is considering using its lake as a source of power. The average temperature difference from the top to the bottom is \(15^{\circ} \mathrm{C},\) and the average surface temperature is \(22^{\circ} \mathrm{C} .\) (a) Assuming that the town can set up a reversible engine using the surface and bottom of the lake as heat reservoirs, what would be its efficiency? (b) If the town needs about \(1.0 \times 10^{8} \mathrm{W}\) of power to be supplied by the lake, how many \(\mathrm{m}^{3}\) of water does the heat engine use per second? (c) The surface area of the lake is $8.0 \times 10^{7} \mathrm{m}^{2}$ and the average incident intensity (over \(24 \mathrm{h})\) of the sunlight is \(200 \mathrm{W} / \mathrm{m}^{2} .\) Can the lake supply enough heat to meet the town's energy needs with this method?

To calculate the efficiency of the reversible engine, we will use the Carnot's efficiency formula: Efficiency = \(1 - \frac{T_{cold}}{T_{hot}}\) The average temperature difference between the top and bottom of the lake is 15 degrees Celsius, and the average surface temperature is 22 degrees Celsius, therefore: \(T_{hot} = 22 + 15 = 37^{\circ} \mathrm{C}\) \(T_{cold} = 22^{\circ} \mathrm{C}\) We need to convert these temperatures to Kelvin: \(T_{hot} = 37 + 273 = 310 \mathrm{K}\) \(T_{cold} = 22 + 273 = 295 \mathrm{K}\) Now, we can plug in these temperatures into the Carnot efficiency formula: Efficiency = \(1 - \frac{295}{310} = 1 - 0.9516 = 0.0484\) Therefore, the efficiency of the reversible engine is approximately 4.84%.

To find the volume of water processed per second, we first need to find the heat input (Q) by using the efficiency formula: Power = Efficiency * Heat Input We are given that the power required by the town is \(1.0 \times 10^8 \mathrm{W}\), and we calculated the efficiency to be 4.84%. \(Q = \frac{Power}{Efficiency} = \frac{1.0 \times 10^8}{0.0484} = 2.066 \times 10^9 \mathrm{W}\) Now we need to find the heat absorbed per unit volume, which can be calculated using the specific heat capacity (c) of water and the temperature difference: Heat Absorbed Per Unit Volume = \(c \times \Delta T\) The specific heat capacity of water is approximately \(4186 \mathrm{J/kg \cdot K}\), and the temperature difference is 15 degrees Celsius. Heat Absorbed Per Unit Volume = \(4186 \times 15 = 62,790 \mathrm{J/m^3}\) Finally, to find the volume flow rate (V) of water processed per second, we can divide the heat input by the heat absorbed per unit volume: \(V = \frac{Q}{Heat Absorbed Per Unit Volume} = \frac{2.066 \times 10^9}{62,790} = 3.29 \times 10^4 \mathrm{m^3/s}\) Thus, the volume of water processed per second is approximately \(3.29 \times 10^4 \mathrm{m^3/s}\).

To determine if the lake can meet the town's energy needs, we need to compare the sunlight energy received by the lake with the required heat input (Q). The average incident intensity of sunlight is given as \(200 \mathrm{W/m^2}\), and the surface area of the lake is given as \(8.0 \times 10^7 \mathrm{m^2}\). Total Sunlight Energy Received = Incident Intensity * Surface Area Total Sunlight Energy Received = \(200 \cdot 8.0 \times 10^7 = 1.6 \times 10^{10} \mathrm{W}\) Now, we need to compare the total sunlight energy received with the required heat input, which we calculated as \(2.066 \times 10^9 \mathrm{W}\). Since the total sunlight energy received (\(1.6 \times 10^{10} \mathrm{W}\)) is greater than the required heat input (\(2.066 \times 10^9 \mathrm{W}\)), the lake can supply enough heat to meet the town's energy needs with this method.