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Chapter 16: Problem 16

A heat engine is to be built to extract energy from the temperature gradient in the ocean. If the surface and deepwater temperatures are \(25^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C},\) respectively, what is the maximum efficiency such an engine can have?

Short Answer

Expert verified
The maximum efficiency of the heat engine is approximately 5.7%.

Step by step solution

01

Identify Temperatures in Kelvin

First, convert the given temperatures from degrees Celsius to Kelvin. The formula to convert degrees Celsius to Kelvin is: \[ T(K) = T(^{\circ}C) + 273.15 \] For the surface temperature: \[ T_1 = 25 + 273.15 = 298.15 \, K \] For the deepwater temperature: \[ T_2 = 8 + 273.15 = 281.15 \, K \]
02

Apply the Carnot Efficiency Formula

The maximum efficiency of a heat engine operating between two temperatures is given by the Carnot efficiency: \[ \eta = 1 - \frac{T_2}{T_1} \] where \( T_1 \) is the temperature of the hot reservoir and \( T_2 \) is the temperature of the cold reservoir.
03

Calculate the Maximum Efficiency

Substitute the Kelvin temperatures into the Carnot efficiency formula: \[ \eta = 1 - \frac{281.15}{298.15} \] Calculate the efficiency: \[ \eta = 1 - 0.943 \approx 0.057 \] Thus, the maximum efficiency is approximately 5.7%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Engine
A heat engine is a device designed to convert thermal energy into mechanical work. It operates on the principle of using heat from a high-temperature source to perform work, while rejecting some of this heat to a low-temperature sink. Here are a few notable characteristics of heat engines:
  • They typically operate on a cycle, continuously converting heat energy to work.
  • The efficiency of a heat engine is determined by the temperatures of the heat source and sink.
  • Examples include car engines, steam turbines in power plants, and even refrigerators (working in reverse).

When analyzing a heat engine utilizing an oceanic temperature gradient, the surface of the ocean acts as the hot reservoir, and the deeper, cooler layers serve as the cold reservoir. Understanding these basic components is key to understanding how the engine generates work from heat.
Temperature Conversion
When working with heat engines, temperature conversion is crucial. This is because equations like the Carnot efficiency formula require temperatures in Kelvin, not Celsius. Here is how you convert Celsius to Kelvin:
  • Use the formula: \[ T(K) = T(^{\circ}C) + 273.15 \]
  • For instance, a surface temperature of \(25^{\circ} \mathrm{C}\) converts to \(298.15 \mathrm{K}\).
  • Similarly, a deepwater temperature of \(8^{\circ} \mathrm{C}\) becomes \(281.15 \mathrm{K}\).

This conversion ensures the correct application of the thermodynamic equations that depend on absolute temperature, reinforcing accuracy in calculations involving heat engines.
Oceanic Temperature Gradient
The oceanic temperature gradient refers to the difference in temperature between shallow and deep ocean waters. This natural gradient can be harnessed to operate a heat engine by using the ocean鈥檚 surface as the high-temperature reservoir and the deeper water as the low-temperature reservoir. Here鈥檚 why this gradient is significant:
  • It provides a sustainable and renewable source of energy.
  • Variations in temperature are consistent, making the system potentially reliable.
  • This temperature difference can power devices using clean, environmentally friendly methods.

Harnessing this gradient effectively could lead to efficient, eco-friendly energy solutions and innovations in harnessing ocean thermal energy.
Maximum Efficiency Calculation
The maximum efficiency of a heat engine using two reservoirs is calculated using the Carnot efficiency formula. This method calculates how efficiently an engine can convert heat into work between two temperatures:The formula is as follows:\[\eta = 1 - \frac{T_2}{T_1}\]where \(T_1\) is the temperature of the hot reservoir and \(T_2\) the cold one.
  • This formula provides the theoretical maximum efficiency; real engines achieve lower efficiencies due to practical limitations like friction and material properties.
  • In our example, substitutions gave us a maximum Carnot efficiency of approximately 5.7%.
  • The closer the temperatures of the two reservoirs, the lower the efficiency 鈥� significant in the context of oceanic gradients.

Thus, understanding maximum efficiency calculations is vital when designing or analyzing engines that exploit such temperature differences.

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Most popular questions from this chapter

A typical doughnut contains approximately 200 food calories (kilocalories), of which about \(80 \%\) is metabolized into heat by your body. Assume that the internal temperature of your body (where digestion occurs) is normally \(37^{\circ} \mathrm{C}\) and that this does not significantly change after you digest the doughnut. By how much does your body's entropy change after you eat a doughnut?

A diesel engine performs \(2200 \mathrm{~J}\) of mechanical work and discards \(4300 \mathrm{~J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

An ice-making machine operates as a Carnot refrigerator. It takes heat from water at \(0.0^{\circ} \mathrm{C}\) and exhausts the heat into a room at \(24.0^{\circ} \mathrm{C}\). Suppose that it converts \(85.0 \mathrm{~kg}\) of water at \(0.0^{\circ} \mathrm{C}\) into ice at \(0.0^{\circ} \mathrm{C}\). (a) How much heat must be removed from the water? (b) How much work energy must be supplied to the refrigerator?

Each cycle, a certain heat engine expels \(250 \mathrm{~J}\) of heat when you put in \(325 \mathrm{~J}\) of heat. Find the efficiency of this engine and the amount of work you get out of the \(325 \mathrm{~J}\) heat input.

A gasoline engine takes in \(1.61 \times 10^{4} \mathrm{~J}\) of heat and delivers \(3700 \mathrm{~J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{~J} / \mathrm{g}\). (a) What is the thermal efficiency of the engine? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?
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