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What would the maximum thermodynamic efficiency be of some heat engine operating between your skin temperature and the ambient environment \(20^{\circ} \mathrm{C}\) cooler than your skin?

Short Answer

Expert verified
Answer: The formula for the maximum thermodynamic efficiency is Carnot efficiency = 1 - ((T_skin - 20) / T_skin), where T_skin is the skin temperature in Celsius.

Step by step solution

01

Convert temperatures to Kelvin

To convert temperatures to Kelvin, add 273.15 to the Celsius values. Let T_skin be the skin temperature in Celsius and T_ambient = T_skin - 20. Then, convert these temperatures to Kelvin using: T_skin_K = T_skin + 273.15 T_ambient_K = T_ambient + 273.15
02

Find the Carnot efficiency

The maximum thermodynamic efficiency of a heat engine is obtained from the Carnot efficiency formula: Carnot efficiency = 1 - (T_cold / T_hot) Where T_cold is the temperature of the cold reservoir (ambient environment) and T_hot is the temperature of the hot reservoir (skin). Plug in the T_ambient_K and T_skin_K values to find the maximum thermodynamic efficiency: Carnot efficiency = 1 - (T_ambient_K / T_skin_K)
03

Simplify the expression

Since both T_skin_K and T_ambient_K have the same constant added (273.15), we can rewrite the expression as Carnot efficiency = 1 - ((T_skin - 20) / T_skin) So, the maximum thermodynamic efficiency for this heat engine operating between skin temperature and ambient environment 20 degrees Celsius cooler than the skin is given by the above formula: Carnot efficiency = 1 - ((T_skin - 20) / T_skin)

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Key Concepts

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

Carnot Efficiency
The concept of Carnot Efficiency is pivotal when discussing the maximum possible efficiency of heat engines. In simple terms, Carnot Efficiency represents the best theoretical performance that a heat engine can reach when converting heat into work. This efficiency is considered the upper limit because it assumes that no energy is lost to friction or other irreversible processes.

Carnot Efficiency is given by the formula: \[\text{Carnot Efficiency} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \]
Where:
  • \(T_{\text{cold}}\) is the temperature of the cold reservoir in Kelvin.
  • \(T_{\text{hot}}\) is the temperature of the hot reservoir in Kelvin.
The relationship highlights that the efficiency only relies on the temperatures of the two reservoirs and is independent of the actual working material of the engine. It shows that to achieve higher efficiency, there needs to be a significant temperature difference between the hot and cold reservoirs.
Heat Engines
Heat engines are devices that convert thermal energy into mechanical work. They operate by taking in energy in the form of heat from a high-temperature source, converting some of it into useful work, and expelling the excess heat to a low-temperature sink. This process is cyclical and continues until the engine stops.

Key points about heat engines include:
  • The efficiency of a heat engine is always less than 100%, due to the second law of thermodynamics, which states that some energy must be lost to the surroundings.
  • Real-world engines cannot be perfectly efficient like an ideal Carnot engine due to inefficiencies such as friction and energy losses.
  • Understanding the limits of these engines helps engineers design better and more efficient machines.
An example of a heat engine could be a car engine, where the chemical energy in fuel is converted into mechanical work to propel the vehicle.
Temperature Conversion
Temperature conversion is an essential skill when dealing with thermodynamics, as calculations often require temperatures to be in Kelvin. Unlike Celsius or Fahrenheit, the Kelvin scale starts from absolute zero, making it ideal for scientific calculations.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature: \[T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15\]This conversion is crucial because thermodynamic equations, such as the ones used for calculating Carnot Efficiency, require temperatures expressed in absolute terms.

Key points to remember for temperature conversion:
  • Kelvin is always positive, as it starts from absolute zero.
  • Conversion is straightforward, involving just a simple addition.
Being comfortable with temperature conversions allows for accurate and meaningful thermodynamic calculations.

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Most popular questions from this chapter

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped a \(5,000 \mathrm{~W}\) heater, what will the internal temperature be if the outside temperature is \(-10^{\circ} \mathrm{C}\) and the heater is running \(100 \%\) of the time?

Let's say you come home from a winter vacation to find your house at \(5^{\circ} \mathrm{C}\) and you want to heat it to \(20^{\circ} \mathrm{C}\). Let's say the house contains: \(500 \mathrm{~kg}\) of air; \(^{62} 1,000 \mathrm{~kg}\) of furniture, books, and other possessions; plus walls and ceiling and floor that amount to \(6,000 \mathrm{~kg}\) of effective \(^{63}\) mass. Using the catch-all specific heat capacity for all of this stuff, how much energy will it take, and how long to heat it up at a rate of \(10 \mathrm{~kW}\) ? Express in useful, intuitive units, and feel free to round, since it's an estimate, anyway.

We can think of wind in the atmosphere as a giant heat engine \(^{67}\) operating between the \(288 \mathrm{~K}\) surface and the top of the troposphere \(^{68}\) at \(230 \mathrm{~K}\). What is the maximum efficiency this heat engine could achieve in converting solar heating into airflow?

Changing from direct electrical heating to a heat pump operating with a COP of 3 means spending one-third the energy for a certain thermal benefit. If a house averages \(30 \mathrm{kWh} /\) day in heating cost through the year using direct electrical heating at a cost of \(\$ 0.15 / \mathrm{kWh}\), how long will it take to recuperate a \(\$ 5,000\) installation cost of a new heat pump?

How much will it cost per day to keep a house at \(20^{\circ} \mathrm{C}\) inside when the external temperature is steady at \(-5^{\circ} \mathrm{C}\) using direct electric heating \(^{65}\) if the house is rated at \(150 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) and electricity costs \(\$ 0.15 / \mathrm{kWh} ?\)

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