Unit 18: Thermodynamics
On January 10, 1962, the town of Pincher Creek experienced a dramatic temperature rise from $-19^\circ\text{C}$ to $+22^\circ\text{C}$ in a period of one hour! After completing this unit, you should be able to explain the phenomenon of chinooks in southwestern Alberta.
Learning Outcomes
By the end of this unit, you should be able to
- define absolute zero.
- state the first law of thermodynamics.
- describe an adiabatic process.
- state the second law of thermodynamics and explain the basic idea behind the heat engine.
- define entropy and explain its relation to order and disorder in the universe.
eText Material
Reading Assignment
Read the following sections in Chapter 18 of the eText:
- 18.1: Absolute Zero
- 18.2: First Law of Thermodynamics
- 18.3: Adiabatic Processes
- 18.4: Meteorology and the First Law
- 18.5: Second Law of Thermodynamics
- 18.6: Energy Tends to Disperse
- 18.7: Entropy
Supplementary learning resources are available on the Mastering Physics learning platform.
Additional Reading
Thermodynamics
When two objects of different temperatures come in contact, thermal energy (or heat) transfers from the warmer object to the cooler object. You can also deliver energy to a gas through work done on it by an external compressive force. Both heat and work are forms of energy that are in transit from one body to another during a given process. The branch of physics that includes the interrelations between work, heat, and internal energy is called thermodynamics.
In studying thermodynamics, you will often use the term system, which refers to all interacting objects under consideration. It is important, however, to distinguish between two types of systems. If energy and material cannot enter or leave the system, then it is called a closed system, or isolated system. Otherwise, it is an open system. Another useful concept is the heat reservoir: a body whose temperature remains practically constant regardless of how much heat flows into or out of it.
First Law of Thermodynamics
The first law of thermodynamics is a restatement of the law of conservation of energy. Suppose a closed system has an internal energy $U$. If an amount $Q$ of heat is added to the system and the work done by the system is $W$, then the change in the internal energy of the system ($\Delta U$) is given by
\begin{equation} Q = \Delta U + W \label{unit18_QUW} \end{equation}
Equation 18.1 is the formal statement of the first law of thermodynamics. Note that when heat is added to the system and work is done by the system, both $Q$ and $W$ are positive. When heat is given off by the system into the environment, $Q$ is negative. Also, if work is done on the system by the environment, then $W$ is negative.
In any thermodynamic process, a system moves from an initial state to a final state, which are defined by variables such as pressure, volume, and temperature. If the system changes from one state to another while the temperature remains constant, the process is referred to as isothermal. In this case, the internal energy of the system does not change (i.e., $\Delta U = 0$). An adiabatic process is one in which no heat enters or leaves the system (i.e., $Q = 0$) during its transition from the initial to the final state.
Second Law of Thermodynamics
The second law of thermodynamics can be stated as follows:
Heat flows naturally from a region of higher temperature to a region of lower temperature.
This statement of the direction of heat flow is one of several alternative forms of the second law of thermodynamics. While it may seem simple and obvious, this law describes one of the most far-reaching and significant discoveries made so far about the nature of our universe.
Heat Engines
A heat engine is a device that converts heat into mechanical work. In each cycle of operation, the engine absorbs heat from the high-temperature ($T_\text{hot}$) reservoir. Part of the absorbed heat is converted to work ($W$), and the remaining heat is discharged to the low-temperature ($T_\text{cold}$) reservoir.
The thermal efficiency of a heat engine is defined as the ratio of the work output and the heat input, or \begin{equation} \text{Efficiency}= \frac{\text{work output}}{\text{heat input}} \end{equation} The formula above suggests that to have a perfect heat engine (100% efficiency), all heat input must be transformed into useful work. However, in 1851, the British mathematician and physicist Lord Kelvin (1824–1907) proposed what is now known as the Kelvin–Planck statement of the second law of thermodynamics, which can be expressed as follows:
It is impossible to design a 100%-efficient heat engine that operates in a cycle and converts all heat input into useful work.
To investigate the efficiency limits of heat engines, French engineer Nicolas Léonard Sadi Carnot (1796–1832) devised an ideal engine in which the working substance was an ideal gas confined to a cylinder with a frictionless piston. He proved mathematically that for such a perfect machine, \begin{equation} \text{Ideal efficiency} = \frac{T_\text{hot} - T_\text{cold}}{T_\text{hot}} = 1 - \frac{T_\text{cold}}{ T_\text{hot}} \end{equation} So, in Carnot’s engine, efficiency is determined solely by the Kelvin temperatures of the hot and cold reservoirs.
Entropy
Carnot’s work established that in the real world, some heat must be wasted in the operation of a heat engine. The total heat in the universe will always be conserved, but the heat that is wasted in the process of doing work will not be available again to do useful work. The arrow of time points only in the direction in which more and more heat is wasted, and even though the total energy remains constant, the quantity of energy available for useful work keeps diminishing.
In 1865, German physicist Rudolf Clausius (1822–1888) introduced the concept of entropy as a measure of the transformation of energy from an available to an unavailable form. According to Clausius, every system has an entropy $S$. When an amount of heat $Q$ is added to a system at a constant temperature $T$, the change in entropy $\Delta S$ of the system is given as \begin{equation} \Delta S = \frac{Q}{T} \end{equation}
Note that the entropy of the system is not defined; Clausius defined only the change in entropy of a system. And it is the change in entropy that is important in the behavior of the universe. The general statement of the second law of thermodynamics in terms of entropy can be worded as follows:
The total entropy of any system and its environment increases as a result of any natural process.
So as entropy keeps increasing with time, more and more energy is degraded into a form that cannot be utilized. In other words, the flow of heat from higher to lower temperatures will continue until the whole cosmos attains the same uniform temperature. The energy will still be there, but it will be incapable of work—a homogeneous system in total equilibrium. The arrow of time is connected with the direction in which the universe unfolds: from order to randomness.
Questions
The following questions are selected from the end of Chapter 18 of the eText. It is important to your learning that you try to answer each question independently before you read through the answer and explanation given.
For questions that ask you to explain, defend, or discuss your answer, the response revealed by the Answer button would earn you only partial marks on a quiz or exam in this course. Use the Answer to help you formulate a complete answer before you select the Explanation button to check your work.
Chapter 18
Question 38
An ocean thermal energy conversion (OTEC) power plant operates on a temperature difference of 4°C deep water and 25°C surface water. Show that the Carnot efficiency of this plant is 7%.
Answer
First, convert the temperatures from degrees Celsius to kelvins: \begin{align} T_\text{cold} &= 4 + 273 = 277\,\text{K} \\[6pt] T_\text{hot} &= 25 + 273 = 298\,\text{K} \\[6pt] \end{align} Then, calculate the ideal (or Carnot) efficiency using Carnot’s equation: \begin{align} \text{Ideal efficiency} &= \frac{T_\text{hot} - T_\text{cold}}{T_\text{hot}} \nonumber\\[6pt] &= \frac{298\,\text{K} - 277\,\text{K}}{298\,\text{K}} \nonumber\\[6pt] &= 0.070 \quad (\text{or}\; 7\%) \end{align}
Chapter 18
Question 46
The temperature of the Sun’s interior is about 15 million degrees. Is it important whether this is degrees Celsius or kelvins? Explain.
Answer
It doesn’t make a big difference which scale is used.
Explanation
If the value of 15 million is the temperature of the Sun’s interior in degrees Celsius, you would write \begin{equation} T_\text{C} = 15{,}000{,}000^\circ\text{C} \end{equation} The corresponding temperature in kelvins is expressed as follows: \begin{align} T_\text{K} &= T_\text{C} + 273 \nonumber\\[6pt] &= 15{,}000{,}273\,\text{K} \end{align} The difference between the two numbers is less than 0.002%, which is insignificant.
Chapter 18
Question 52
Suppose you do 100 J of work in compressing a gas. If 80 J of heat escapes in the process, what is the change in internal energy of the gas?
Answer
It increases by 20 J.
Explanation
Doing work on a system adds to its internal energy, while escaping heat reduces its internal energy. Therefore, the internal energy of the compressed gas increases by \begin{equation} 100\,\text{J} - 80\,\text{J} = 20\,\text{J} \end{equation}
You can also use Equation 18.1 to answer the question. In this case, note that $W = -100\,\text{J}$ and $Q = -80\,\text{J}$.
Chapter 18
Question 56
What is the ultimate source of energy in a hydroelectric power plant? Defend your answer.
Answer
The Sun is considered the ultimate source of energy.
Explanation
The Sun heats water reservoirs such as oceans and lakes, causing water to evaporate. In other words, that solar energy converts into gravitational potential energy stored in the clouds and the large artificial lakes behind dams. The gravitational potential energy then transforms into rotational mechanical energy in the turbines of the hydroelectric power plant and then into electric energy.
Chapter 18
Question 57
Why is it advantageous to use steam that is as hot as possible in a steam-driven turbine?
Answer
It results in greater efficiency.
Explanation
The higher the temperature ($T_\text{hot}$) of the steam compared with the exhaust temperature ($T_\text{cold}$), the greater the efficiency of the engine. In other words, a greater fraction of the input energy can be converted to useful work.
Chapter 18
Question 64
A refrigerator moves heat from cold to warm. Why does this not violate the second law of thermodynamics?
Answer
Because work is done by an external agent.
Explanation
Heat does not flow spontaneously from the interior of the refrigerator to the external environment. In this case, a heat engine (the compressor) does external work, powered by electric energy, to pump heat from the cold chamber to the warmer environment.
Chapter 18
Question 65
If you squeeze an air-filled balloon and no heat escapes, what happens to the internal energy of the gas in the balloon?
Answer
It increases.
Explanation
The work done by your muscles while squeezing the balloon adds to the internal energy of the gas inside. Since no heat escapes (adiabatic process), the internal energy (and temperature) of the gas inside the balloon increases.
Chapter 18
Question 69
According to the second law of thermodynamics, is the universe moving to a more ordered state or to a more disordered state?
Answer
To a more disordered state.
Explanation
According to the second law of thermodynamics, the net entropy (or amount of disorder) in the universe increases with time. As heat continues to transfer from warmer to colder objects, the universe tends toward a uniform distribution of temperature. As a result, the amount of available useful energy gradually decreases with time.
Chapter 18
Question 71
The ocean possesses enormous numbers of molecules, all with kinetic energy. Can this energy be extracted and used as a power source? Defend your answer.
Answer
Yes, but not very efficiently. The challenge in extracting energy from the ocean is finding a cold reservoir with a lower temperature. The temperature difference between the water at the top and bottom of the ocean can be used, but with low efficiency (see Question 38).
Chapter 18
Question 73
Water evaporates from a salt solution and leaves behind salt crystals that have a higher degree of molecular order than the more randomly moving molecules in the saltwater. Has the entropy principle been violated? Why or why not?
Hint
Also consider the entropy change in the water when it evaporates.
Answer
No, the entropy principle is still valid. The degree of molecular order of the salt molecules increases as they arrange into salt crystals; however, the process generates greater disorder in the water molecules that turn into steam. As a result, the overall amount of disorder (or entropy) of the system increases.
Chapter 18
Question 80
Under what theoretical conditions would a heat engine be 100% efficient?
Answer
When the sink is at absolute zero.
Explanation
Perfect (or 100%) efficiency can be expressed as follows: \begin{align} \text{Ideal efficiency} = \frac{T_\text{hot} - T_\text{cold}}{T_\text{hot}} &= 1 \nonumber\\[6pt] \Rightarrow\quad 1 - \frac{T_\text{cold}}{T_\text{hot}} &= 1 \nonumber\\[6pt] \Rightarrow\quad \frac{T_\text{cold}}{T_\text{hot}} &= 0 \end{align} This is achieved when $T_\text{cold} = 0\,\text{K}$ or when $T_\text{hot}$ is infinitely large.
Chapter 18
Question 81
Heat always flows spontaneously from an object with a higher temperature to an object with a lower temperature. Is this the same thing as saying that heat always flows from an object with a greater internal energy to one with a lower internal energy? Discuss why or why not.
Hint
As a analogy, think about the difference between the height of a container and the amount of water in it.
Answer
No, it is not the same. The internal energy of a lake is much greater than the internal energy of a glowing piece of charcoal. However, if you throw the charcoal into the lake, heat transfers from the object with smaller internal energy (the charcoal) to the object with greater internal energy (the lake).
Exercises
Spend some time completing the following exercises to test your understanding of the main concepts in Chapter 18 and increase your efficiency in answering exam questions.
End-of-Chapter Practice Questions
Answer questions 5, 11, 35, 39, 41, 47, 51, 59, 61, 75, 77, and 83 in Chapter 18 of the eText. If you require assistance, please contact your tutor. The answers are provided at the end of the eText.
Quiz 4
Before moving on to finish your course project and request to write your final exam, complete Quiz 4, which covers Units 15–18. For more information and to take the quiz, see the Quizzes section on the course home page. This quiz is worth 10% of your course grade.