A town is planning on using the water flowing through a river at a rate of \(5.0 \times 10^{6} \mathrm{kg} / \mathrm{s}\) to carry away the heat from a new power plant. Environmental studies indicate that the temperature of the river should only increase by \(0.50^{\circ} \mathrm{C} .\) The maximum design efficiency for this plant is \(30.0 \% .\) What is the maximum possible power this plant can produce?

First, we need to determine the amount of heat that the water can carry away. To do so, we need to use the specific heat capacity formula: \(Q = mcΔT\) where \(Q\) is the heat energy transferred, \(m\) is the mass flow rate of the water (in kg/s), \(c\) is the specific heat capacity of water, which is approximately \(4.186 \times 10^{3} \mathrm{J/kg \cdot K}\), and \(ΔT\) is the temperature change (in Kelvin or Celsius). We are given the mass flow rate \(m = 5.0 \times 10^6 \mathrm{kg/s}\) and the temperature change \(ΔT = 0.50^\circ \mathrm{C}\). Plugging these values into the formula, we can find the heat energy transferred: \(Q = (5.0 \times 10^6 \mathrm{kg/s})(4.186 \times 10^3 \mathrm{J/kg\cdot K})(0.50^\circ \mathrm{C})\)

Now, we can compute the heat energy transferred per second: \(Q = 5.0 \times 10^6 \times 4.186 \times 10^3 \times 0.50 \approx 1.0465 \times 10^{10} \mathrm{J/s}\)

Finally, we use the given efficiency to determine the maximum possible power output of the plant: Power \(= \mathrm{Efficiency} \times \mathrm{Heat \ energy \ transferred \ per \ second}\) We are given an efficiency of \(30.0\%\), which we will convert to a decimal value: \(0.30 = \frac{30.0}{100}\) Now, we can find the maximum possible power: Power \(= 0.30 \times 1.0465 \times 10^{10} \approx 3.1395 \times 10^{9} \mathrm{W}\) Therefore, the maximum possible power this plant can produce is approximately \(3.14 \times 10^9 \mathrm{W}\).