Question

In tropical climates, the water near the surface of the ocean remains warm throughout the year as a result of solar energy absorption. In the deeper parts of the ocean, however, the water remains at a relatively low temperature since the sun's rays cannot penetrate very far. It is proposed to take advantage of this temperature difference and construct a power plant that will absorb heat from the warm water near the surface and reject the waste heat to the cold water a few hundred meters below. Determine the maximum thermal efficiency of such a plant if the water temperatures at the two respective locations are 24 and \(3^{\circ} \mathrm{C}\)

Short answer

Answer: The maximum thermal efficiency of such a power plant is 7.08%.

Step-by-step solution

Convert temperatures to Kelvin

First, we should convert the given temperatures from Celsius to Kelvin. Remember that to convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature. Therefore: Warm water temperature, \(T_H = 24 + 273.15 = 297.15 \,\text{K}\) Cold water temperature, \(T_C = 3 + 273.15 = 276.15 \,\text{K}\)

Calculate Carnot efficiency

The maximum thermal efficiency of a heat engine (in this case, the power plant) operating between two temperatures is given by the Carnot efficiency formula: \(\eta_{max} = 1 - \frac{T_C}{T_H}\) Now, we can plug in the values of \(T_H\) and \(T_C\): \(\eta_{max} = 1 - \frac{276.15\,\text{K}}{297.15\,\text{K}}\)

Find the maximum thermal efficiency

Now, we simply calculate the result: \(\eta_{max} = 1 - \frac{276.15}{297.15} = 1 - 0.9292 = 0.0708\) To express this as a percentage, we multiply by 100: Maximum thermal efficiency = \(0.0708 \times 100 = 7.08 \%\) Thus, the maximum thermal efficiency of such a power plant operating between the warm surface water and cold deep water in tropical climates is 7.08%.