Suppose that 100. J of energy is taken from a hot source at \(300 .{ }^{\circ} \mathrm{C}\), passes through a turbine that converts some of the energy into work, and then releases the rest of the energy as heat into a cold sink at \(20 .{ }^{\circ} \mathrm{C}\). What is the maximum amount of work that can be produced by this engine if overall it is to operate spontaneously? What is the efficiency of the engine, with work done divided by heat supplied expressed as a percentage? How could the efficiency be increased?

Short Answer

Expert verified
The maximum work is approximately 48.81 J, and the Carnot efficiency is approximately 48.81%. Efficiency could be increased by raising the temperature of the hot source or lowering the temperature of the cold sink.

Step by step solution

01

Calculate the Carnot Efficiency

The Carnot efficiency is the theoretical maximum efficiency of a heat engine operating between two temperatures. It is calculated using the formula \(\eta_{Carnot} = 1 - \frac{T_c}{T_h}\), where \(T_c\) and \(T_h\) are the absolute temperatures of the cold and hot reservoirs respectively in Kelvin. First, convert the temperatures from Celsius to Kelvin by adding 273.15 to each. Hot source: \(300 ^{\circ}C = 300 + 273.15 = 573.15 K\) and Cold sink: \(20 ^{\circ}C = 20 + 273.15 = 293.15 K\). Then apply the formula: \(\eta_{Carnot} = 1 - \frac{293.15}{573.15}\).
02

Convert the Efficiencies into Percentages

The Carnot efficiency needs to be expressed as a percentage. Do this by multiplying the efficiency by 100. If \(\eta_{Carnot}\) is the efficiency as a decimal, the efficiency percentage is given by \(\eta_{Carnot} \times 100\%\).
03

Calculate the Maximum Work Output

The maximum amount of work that can be produced is the total energy multiplied by the Carnot efficiency. With \(100 J\) of energy supplied, the work (\(W_{max}\)) is \(W_{max} = \eta_{Carnot} \times 100 J\).
04

Answer the Question on Efficiency Increase

To increase the efficiency of the heat engine, either the temperature of the hot source could be increased or the temperature of the cold sink could be decreased, as this will increase the temperature difference in the Carnot efficiency formula, given other factors remain constant.

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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 the relationships between heat and other forms of energy. It is primarily concerned with the concepts of energy transfer, conversion, and the laws that govern these processes. In the context of our exercise, thermodynamics provides the foundational principles that define the operation of a heat engine, which is a device that converts thermal energy into mechanical work.

Within thermodynamics, there are several laws, but two particularly relevant to our exercise are the first law, which states that energy cannot be created or destroyed, only transformed, and the second law, which addresses the direction of energy transfer and the quality of energy. The exercise relates closely to these laws by demonstrating how heat energy is converted into useful work and highlighting constraints on efficiency due to the temperatures involved.
Heat Engines
A heat engine is a system that converts heat or thermal energy into mechanical energy, or work. The performance of a heat engine is determined by its efficiency, which is the ratio of work output to heat input. In the textbook exercise, the engine takes in 100 joules of heat energy from a hot source and then does work by moving a turbine before discharging lower-quality heat to a cold sink. This is a classic example of a heat engine cycle.

Heat engines can be found in many applications such as car engines and power plants. They operate between two temperature reservoirs, absorbing heat from the higher-temperature source and releasing some heat to the lower-temperature sink. The engine's ability to do work stems from this temperature difference. The Carnot efficiency sets a theoretical limit on the efficiency that real engines can achieve, as dictated by the second law of thermodynamics.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In our exercise, the engine converts thermal energy (heat) into mechanical energy (work). The conversion of energy is governed by the laws of thermodynamics, which dictate that while energy in a closed system remains constant, its form can change, and the quality of energy can degrade.

During the conversion process, some energy is always lost as waste heat, which is why no practical engine can reach 100% efficiency. The Carnot engine represents an idealized model where the conversion process is reversible and no energy is lost to friction or other processes. It sets the upper limit of efficiency for any real heat engine, although real engines will invariably fall short of this theoretical efficiency due to real-world inefficiencies.
Temperature Conversion
Temperature conversion is necessary to work with thermodynamic equations, which require temperatures to be in an absolute scale. In our step-by-step solution, we converted temperatures from Celsius to Kelvin because the Kelvin scale sets its zero point at absolute zero, the theoretical point where there is no thermal energy in a system.

To convert Celsius to Kelvin, one simply adds 273.15 to the Celsius temperature. This is important because the efficiency calculations for a Carnot engine depend on the absolute temperatures of the hot and cold reservoirs. Using absolute temperatures allows for consistent thermodynamic calculations, as it is a true representation of thermal energy content.
Second Law of Thermodynamics
The second law of thermodynamics introduces the concept of entropy, stating that for any spontaneous process, the total entropy of a system and its surroundings must increase. In simpler terms, it means energy tends to spread out or disperse if not hindered. For a heat engine like the one described in our exercise, this law implies that some energy will always be lost to the surroundings and it is impossible to convert all the absorbed heat into work.

This law also tells us why we cannot have a 100% efficient heat engine. As per the Carnot principle, the maximum efficiency depends on the temperature of the hot and cold reservoirs and it shows that to improve the efficiency of an engine, one can increase the temperature of the hot source or decrease that of the cold sink thus, enhancing the temperature difference and, consequently, the potential for work output. This aligns directly with the question of how efficiency could be increased in our original problem.

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Most popular questions from this chapter

Propose the argument that, for any liquid at atmospheric pressure (that is, a liquid that boils above room temperature when the external pressure is \(1 \mathrm{~atm}\) ), the numerical value of \(\Delta H_{\text {vap }}\) in joules per mole is greater than the numerical value of \(\Delta S_{\text {vap }}\) in joules per kelvin per mole. (Explain and justify each step and any assumptions.)

A human body generates heat at the rate of about 100 . W \(\left(1 \mathrm{~W}=1 \mathrm{~J} \cdot \mathrm{s}^{-1}\right)\). (a) At what rate does your body heat generate entropy in your surroundings, taken to be at 20 . \(^{\circ} \mathrm{C}\) ? (b) How much entropy do you generate each day? (c) Would the entropy generated be greater or less if you were in a room kept at \(30 .{ }^{\circ} \mathrm{C}\) ? Explain your answer.

The standard entropy of vaporization of acetone is approximately \(85 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\) at its boiling point. (a) Estimate the standard enthalpy of vaporization of acetone at its normal boiling point of \(56.2^{\circ} \mathrm{C}\). (b) What is the entropy change of the surroundings when 10. \(\mathrm{g}\) of acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), condenses at its normal boiling point?

Three isomeric alkenes have the formula \(\mathrm{C}_{4} \mathrm{H}_{8}\) (see the following table). (a) Draw Lewis structures of these compounds. (b) Calculate \(\Delta G^{\circ}, \Delta H^{\circ}\), and \(\Delta S^{\circ}\) for the three reactions that interconvert each pair of compounds. (c) Which isomer is the most stable? (d) Rank the isomers in order of decreasing \(S_{m}{ }^{\circ}\). \(\begin{array}{lrc} \text { Compound } & \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\right) & \Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\right) \\ \hline \text { 2-methylpropene } & -16.90 & +58.07 \\ \text { cis-2-butene } & -6.99 & +65.86 \\ \text { trans-2-butene } & -11.17 & +62.97 \\ \hline \end{array}\)

Calculate the entropy change associated with the isothermal expansion of \(5.25 \mathrm{~mol}\) of ideal gas atoms from \(24.252 \mathrm{~L}\) to \(34.058 \mathrm{~L}\).

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