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20.14. A Carmot engine is operated between two heat reservoirs at temperatures of 520 \(\mathrm{K}\) and 300 \(\mathrm{K}\) (a) If the engine receives 6.45 \(\mathrm{kJ}\) of heat energy from the reservoir at 520 \(\mathrm{K}\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(\mathrm{K}\) ? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

Short Answer

Expert verified
(a) 3721.15 J (b) 2728.85 J (c) 42.31%

Step by step solution

01

Understand the Carnot Engine Basics

A Carnot engine is a theoretical device that operates on a reversible Carnot cycle and has two heat reservoirs: one hot (at temperature \(T_h\)) and one cold (at temperature \(T_c\)). The efficiency of a Carnot engine depends on these temperatures and is given by \(\eta = 1 - \frac{T_c}{T_h}\), where \(T_h\) and \(T_c\) are in Kelvin.
02

Calculate Thermal Efficiency

Given the temperatures \(T_h = 520\, \mathrm{K}\) and \(T_c = 300\, \mathrm{K}\), the thermal efficiency \(\eta\) is calculated as follows:\[ \eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{520} = 1 - 0.5769 \approx 0.4231 \text{ or } 42.31\% \]
03

Relate Heat Absorbed and Rejected

For a Carnot engine, the relationship between the heat absorbed \(Q_h\), the heat rejected \(Q_c\), and the work done \(W\) is \(Q_h = W + Q_c\). We can also use \(Q_c = Q_h \left( \frac{T_c}{T_h} \right)\) to find the heat discarded.
04

Find Heat Discarded to the Cold Reservoir

Given that \(Q_h = 6.45\, \mathrm{kJ} = 6450\, \mathrm{J}\), calculate \(Q_c\) using:\[ Q_c = Q_h \left( \frac{T_c}{T_h} \right) = 6450 \times \frac{300}{520} \approx 3721.15\, \mathrm{J} \]
05

Determine Mechanical Work Done

Using the relationship \(W = Q_h - Q_c\), we find the work done by the engine:\[ W = 6450 - 3721.15 \approx 2728.85\, \mathrm{J} \]
06

Summarize Results

(a) The engine discards approximately 3721.15 J of heat per cycle to the cold reservoir. (b) The mechanical work performed by the engine per cycle is approximately 2728.85 J. (c) The thermal efficiency of the engine is about 42.31%.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and the forms of energy. It helps us understand how energy is transferred and transformed in different systems. The Carnot engine is a central concept in thermodynamics because it represents the most efficient possible engine. It operates on the Carnot cycle, which consists of four reversible processes: two isothermal and two adiabatic. This cycle allows the engine to draw heat from a hot reservoir, perform work, and release some heat to a cold reservoir. The study of thermodynamics also involves understanding the laws of thermodynamics:
  • The Zeroth Law, which helps define temperature.
  • The First Law, emphasizing energy conservation.
  • The Second Law, which introduces the concept of entropy.
  • The Third Law, which deals with absolute zero temperature.
Understanding these laws helps us predict how energy flows in engines like the Carnot engine, making them incredibly important in physics and engineering.
Heat Reservoirs
A heat reservoir is a system that supplies or absorbs finite amounts of heat without any change in temperature. In the context of a Carnot engine, there are two heat reservoirs:
  • The hot reservoir at a higher temperature, which provides heat energy.
  • The cold reservoir at a lower temperature, where the engine discards heat energy.
These two reservoirs are crucial for the operation of a Carnot engine because they allow the engine to maintain a temperature difference, enabling it to perform work. Typically, the greater the temperature difference between the reservoirs, the more efficient the engine can be. Heat reservoirs are assumed to remain unchanged during the heat exchange process, making them ideal systems in thermodynamics. The interplay between these reservoirs is what makes the thermodynamic cycle possible.
Thermal Efficiency
Thermal efficiency (\( \eta \)) is a measure of how well an engine converts heat from the hot reservoir into useful work. In a Carnot engine, thermal efficiency is given by the formula:\[\eta = 1 - \frac{T_c}{T_h}\]where \( T_c \) is the temperature of the cold reservoir, and \( T_h \) is the temperature of the hot reservoir, both measured in Kelvin. This equation demonstrates that the thermal efficiency depends solely on the temperatures of the reservoirs. The closer the temperature of the cold reservoir is to absolute zero, or the higher the temperature difference between the two reservoirs, the closer an engine can approach 100% efficiency. However, real-world factors such as friction and material limitations prevent any actual engine from achieving Carnot efficiency. For our Carnot engine, with temperatures of 520 K and 300 K, the thermal efficiency is approximately 42.31%. This means that only about 42.31% of the heat absorbed is converted into mechanical work.
Mechanical Work
Mechanical work in the context of a Carnot engine refers to the useful energy output that can perform tasks, such as moving a piston or activating a turbine. In a Carnot cycle, the work output (\( W \)) is achieved by converting part of the heat energy absorbed from the hot reservoir. Using the relationship:\[ W = Q_h - Q_c \]where \( Q_h \) is the heat absorbed from the hot reservoir, and \( Q_c \) is the heat rejected to the cold reservoir, we find the mechanical work done.In the given problem, with \( Q_h = 6450 \) J and \( Q_c = 3721.15 \) J, the work done per cycle is approximately 2728.85 J. This value represents the portion of heat energy that has been transformed into mechanical motion, useful for doing work on other systems or mechanisms. Understanding how this transformation takes place helps us design engines and machines that are efficient and effective in performing their intended tasks.

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Most popular questions from this chapter

20.16. An ice-making machine operates in a Camot cycle. It takes heat from water at \(0.0^{\circ} \mathrm{C}\) and rejects heat to a room at \(24.0^{\circ} \mathrm{C}\) . Suppose that 85.0 \(\mathrm{kg}\) of water at \(0.0^{\circ} \mathrm{C}\) are converted to ice at \(0.0^{\circ} \mathrm{C}\) . (a) How much heat is discharged into the room? (b) How much energy must be supplied to the device?

20\. 07. What compression ratio \(r\) must an Otto cycle have to achieve an ideal efficiency of 65.0\(\%\) if \(\gamma=1.40 ?\)

20.38. You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 \(\mathrm{atm}\) . With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 kg block of ice originally at \(0.0^{\circ} \mathrm{C}\) , using only the beat rejected by the engine?

20.11. A window air-conditioner unit absorbs \(9.80 \times 10^{4} \mathrm{J}\) of heat per minute from the room being cooled and in the same time period deposits \(1.44 \times 10^{5} \mathrm{J}\) of heat into the outside air (a) What is the power consumption of the unit in watts? (b) What is the energy efficiency rating of the unit?

20.15. A Carnot engine has an efficiency of 59\% and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature ( \(20.0^{\circ} \mathrm{C} )\) . What is the temperature of its heat source?

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