91Ó°ÊÓ

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 \(3.14 \times 10^9 \mathrm{W}\).