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 68 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, we can use the formula for calculating the maximum efficiency of a heat engine. The formula is:
Efficiency = 1 - (T_cold / T_hot)
Where T_cold is the temperature of the cold reservoir and T_hot is the temperature of the hot reservoir. In this case, the cold reservoir is the water temperature at a depth of about 1 km, which is approximately 5.0 ̊C (278.15 K). The hot reservoir is the surface temperature of the ocean, which is approximately 20 ̊C (293.15 K).
Given that the power output of the plant is 68 MW, we can convert it to joules per second (J/s) by multiplying by 10^6:
Power output = 68 MW = 68 × 10^6 J/s
Now, we can rearrange the formula for efficiency to solve for the energy taken from the hot reservoir:
Efficiency = 1 - (T_cold / T_hot)
Energy taken from hot reservoir = Efficiency × Power output
Plugging in the values:
Efficiency = 1 - (278.15 K / 293.15 K)
Energy taken from hot reservoir = (1 - (278.15 K / 293.15 K)) × 68 × 10^6 J/s
Calculating this expression will give us the amount of energy taken from the warm reservoir in joules per second (J/s).