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(III) A real heat engine working between heat reservoirs at 970 K and 650 K produces 550 J of work per cycle for a heat input of 2500 J. (\(a\)) Compare the efficiency of this real engine to that of an ideal (Carnot) engine. (\(b\)) Calculate the total entropy change of the universe per cycle of the real engine, and (\(c\)) also if the engine is ideal (Carnot).

Short Answer

Expert verified
(a) Real engine efficiency: 22%, Carnot efficiency: 32.99%. (b) Total entropy change (real engine): 0.423 J/K. (c) Entropy change (Carnot engine): 0 J/K.

Step by step solution

01

Understanding Efficiency

The efficiency of a heat engine is calculated using the formula \( \eta = \frac{W}{Q_{in}} \), where \( W \) is the work done and \( Q_{in} \) is the heat input. For the real engine, with \( W = 550 \text{ J} \) and \( Q_{in} = 2500 \text{ J} \), we calculate: \( \eta_{\text{real}} = \frac{550}{2500} = 0.22 \text{ or } 22\% \).
02

Calculate Carnot Efficiency

The efficiency of an ideal Carnot engine working between two temperature reservoirs is given by \( \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \). Using \( T_H = 970 \text{ K} \) and \( T_C = 650 \text{ K} \), \( \eta_{\text{Carnot}} = 1 - \frac{650}{970} = 0.3299 \text{ or } 32.99\% \).
03

Efficiency Comparison

Comparing the efficiencies, the real engine has an efficiency of 22% while the Carnot engine has an efficiency of 32.99%. The real engine is less efficient than the Carnot engine.
04

Calculate Entropy Change for the Real Engine

The change in entropy for the universe due to the real engine can be calculated using the formula \( \Delta S = \Delta S_H + \Delta S_C \), where \( \Delta S_H = -\frac{Q_H}{T_H} \) and \( \Delta S_C = \frac{Q_C}{T_C} \). First find \( Q_C \) with energy balance: \( Q_C = Q_{in} - W = 2500 \text{ J} - 550 \text{ J} = 1950 \text{ J} \). Thus, \( \Delta S_H = -\frac{2500}{970} \) and \( \Delta S_C = \frac{1950}{650} \). Calculate separately and sum the entropies.
05

Calculate Entropy Change for the Carnot Engine

A Carnot engine is reversible and thus has no net change in entropy of the universe, so \( \Delta S_{\text{universe, Carnot}} = 0 \).
06

Numerical Entropy Calculations

For the real engine, \( \Delta S_H = -2.577 \text{ J/K} \) and \( \Delta S_C = 3.000 \text{ J/K} \). Total entropy change is \( \Delta S_{\text{universe, real}} = 3.000 - 2.577 = 0.423 \text{ J/K} \).

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Key Concepts

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

Efficiency of Heat Engines
Heat engines are devices that convert thermal energy into mechanical work. The efficiency of a heat engine is a measure of how well the engine converts the input heat energy into useful work. It is defined by the formula \[ \eta = \frac{W}{Q_{in}} \] where \( W \) is the work done by the engine and \( Q_{in} \) is the heat input. In our example, the real engine showed an efficiency of 22%. This means it converts 22% of the input heat energy into work. The efficiency calculation is crucial when evaluating thermal engines because it helps us understand how effectively energy resources are being utilized. An engine's efficiency can never be 100% due to losses, such as friction and energy dissipations. The closer the efficiency is to 100%, the better the engine is at converting heat into work. The study of engine efficiencies allows engineers to design systems that consume less fuel and have less environmental impact.
Carnot Engine
The Carnot engine is an idealized model that represents the maximum possible efficiency any heat engine can achieve operating between two thermal reservoirs. This theoretical engine sets an upper limit on efficiency by avoiding all forms of energy loss. The efficiency of a Carnot engine is given by:\[ \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \]where \( T_H \) is the temperature of the hot reservoir and \( T_C \) is the temperature of the cold reservoir, both in Kelvin.In our problem, the Carnot engine efficiency came out to be 32.99%, illustrating the performance of an ideal engine operating between the same reservoirs as the real engine. While real engines like those used in cars and power plants experience losses, understanding the Carnot efficiency gives us a benchmark. Knowing this ideal efficiency helps engineers pinpoint areas where real engines can be improved and how close they are to the maximum potential efficiency. Despite being theoretical, the Carnot engine remains a critical concept in thermodynamics and helps identify the efficiency of real engines.
Entropy Change
Entropy is a term in thermodynamics that measures the degree of disorder or randomness in a system. When heat engines operate, they cause changes in entropy, both within themselves and in their surroundings. For the real heat engine in our problem, the total entropy change in the universe was calculated as 0.423 J/K per cycle. Entropy change is split into contributions from the hot and cold reservoirs:
  • \( \Delta S_H = -\frac{Q_H}{T_H} \)
  • \( \Delta S_C = \frac{Q_C}{T_C} \)
In a real engine, there is always some increase in entropy, indicating irreversibilities such as friction. For ideal engines, like the Carnot engine, the process is reversible, resulting in zero net change in entropy of the universe.Understanding these concepts can show us the inefficiencies and help improve engine designs. Following the second law of thermodynamics, no process can decrease the entropy of the universe. This principle is used to evaluate the environmental impact of energy conversion processes, promoting designs that align closer to reversible, low-entropy processes.

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Most popular questions from this chapter

(II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a \(PV\) diagram, including numbers and labels for the axes.

(II) What volume of water at 0\(^\circ\)C can a freezer make into ice cubes in 1.0 h, if the coefficient of performance of the cooling unit is 6.0 and the power input is 1.2 kilowatt?

(II) A heat engine uses a heat source at 580\(^\circ\)C and has an ideal (Carnot) efficiency of 22\(\%\). To increase the ideal efficiency to 42\(\%\), what must be the temperature of the heat source?

(\(a\)) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750\(^\circ\)C and 440\(^\circ\)C, and of the second 415\(^\circ\)C and 270\(^\circ\)C. If the heat of combustion of coal is \(2.8 \times 10^7\) J/kg, at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65\(\%\) of the ideal (Carnot) efficiency. (\(b\)) Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 4.5 C\(^\circ\), estimate how much water must pass through the plant per hour.

(I) A heat engine does 9200 J of work per cycle while absorbing 25.0 kcal of heat from a high-temperature reservoir. What is the efficiency of this engine?

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