As an environmentally friendly idea, some have proposed using the temperature gradient of the ocean to generate electricity. This system would use the temperature gradient between the surface of the ocean (at about 200 C) and the water temperature at a depth of about 1 km (at about 5.0 0C). Assume the power output of such a plant is 76 MW, how much energy is taken from the warm reservoir? Assumee the plant operates at maximum efficiency. Express your answer in J/s
To calculate the amount of energy taken from the warm reservoir in the ocean, we can use the formula:
Energy = Power × Time
Given that the power output of the plant is 76 MW (megawatts), we can convert it to watts by using the following conversion:
1 MW = 1,000,000 watts
So, 76 MW = 76,000,000 watts
Assuming the plant operates at maximum efficiency, the amount of energy taken from the warm reservoir over a certain period of time will be equal to the power output multiplied by that time.
Let's calculate the amount of energy taken per second:
Energy = 76,000,000 watts × 1 second
To simplify the calculation, let's convert the temperature gradient from Celsius to Kelvin since the Kelvin scale is absolute:
Temperature at the surface of the ocean (T1) = 200 °C = 200 + 273.15 K = 473.15 K
Temperature at a depth of 1 km (T2) = 5.0 °C = 5.0 + 273.15 K = 278.15 K
Now, we can calculate the temperature difference (ΔT) between the warm reservoir and the cold reservoir:
ΔT = T1 - T2
ΔT = 473.15 K - 278.15 K
ΔT = 195 K
To find the energy taken from the warm reservoir (Q), use the formula:
Q = Energy = Efficiency × (ΔT)
= Power × Time
= 76,000,000 watts × 1 second
= 76,000,000 joules
Therefore, the amount of energy taken from the warm reservoir is 76,000,000 joules per second (J/s).