A heat engine uses blackbody radiation as its operating substance. The equation of state for blackbody radiation is \(P=1 / 3 a T^{4}\) and the internal energy is \(U=a V T^{4}\), where \(a=7.566 \times 10^{-16}\) \(\mathrm{J} /\left(\mathrm{m}^{3} \mathrm{~K}^{4}\right)\) is Stefan's constant, \(P\) is pressure, \(T\) is temperature, and \(V\) is volume. The engine cycle consists of three steps. Process \(1 \rightarrow 2\) is an expansion at constant pressure \(P_{1}=P_{2} .\) Process \(2 \rightarrow 3\) is a decrease in pressure from \(P_{2}\) to \(P_{3}\) at constant volume \(V_{2}=V_{3}\). Process \(3 \rightarrow 1\) is an adiabatic contraction from volume \(V_{3}\) to \(V_{1}\). Assume that \(P_{1}=3.375 P_{3}, T_{1}=2000 \mathrm{~K}\), and \(V_{1}=10^{-3} \mathrm{~m}^{3}\). (a) Express \(V_{2}\) in terms of \(V_{1}\) and \(T_{1}=T_{2}\) in terms of \(T_{3}\) (b) Compute the work done during each part of the cycle. (c) Compute the heat absorbed during each part of the cycle. (d) What is the efficiency of this heat engine (get a number)? (e) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures.

Short Answer

Expert verified
V2 = V1, T2 = T1, Work done is zero through all processes, no heat absorbed, and efficiency is 0%. Carnot efficiency is 38.4%.

Step by step solution

01

Title - Determine Volume V2 Using V1 and P1

Given that the process from 1 to 2 is at constant pressure and the blackbody radiation equation of state is given by \(P=1 / 3 a T^{4}\). Since pressure is constant, the volume relationship in terms of temperature and Stefan's constant can be derived as follows: \(V_{2} = V_{1} \frac{T_{1}^4}{T_{2}^4}\). Since it’s given that \(T_1 = T_2\), therefore \(V_2 = V_1\).
02

Title - Determine T2 in terms of T3

Given \(P_{1}=P_{2}\), use the blackbody radiation equation \(P=1 / 3 a T^{4}\) to get \(T_{2} = T_{1}\) because it’s an isobaric process and pressure remains constant. Since \(P_{3} = \frac{P_{1}}{3.375} = 0.296 \times P_{1}\), we find \(T_{3}\rightarrow P_{3} = 1 / 3 a T_{3}^{4}\). Thus, \(T_3 = T_1 \times \bigg(\frac{P_3}{P_1}\bigg)^{1/4}\).
03

Title - Compute Work Done During Each Part of the Cycle

For Process 1 \(\rightarrow\) 2 (Constant pressure expansion): \(W_{1 \rightarrow 2} = P_{1}(V_{2} - V_{1})\). Substituting the given values, \(V_{2} = V_{1}, P_{1} = 3.375P_{3}\), the work done, \( W_{1 \rightarrow 2} = 0\).For Process 2 \(\rightarrow\) 3 (Constant volume cooling): \(W_{2 \rightarrow 3} = 0\).For Process 3 \(\rightarrow\) 1 (Adiabatic contraction): Adiabatic process work formula is \(W_{3 \rightarrow 1} = U_{3} - U_{1}\), \(W_{3 \rightarrow 1} = a (V_{3}T_{3}^{4} - V_{1}T_{1}^{4})\).
04

Title - Compute Heat Absorbed During Each Part of the Cycle

During Process 1 \(\rightarrow\) 2: \(Q_{1 \rightarrow 2} = \frac{4}{3}a(T_1^3) \times (V_2 - V_1) = 0\).For Process 2 \(\rightarrow\) 3, since the volume is constant, no heat is absorbed, \(Q_{2 \rightarrow 3} = 0\).For Process 3 \(\rightarrow\) 1: No heat absorbed as it's adiabatic, \(Q_{3 \rightarrow 1}= 0\).
05

Title - Calculate the Efficiency of the Heat Engine

Efficiency \(\text{Eff} = \frac{\text{Net Work Output}}{\text{Heat Input}}\). For all processes, \(W_{net} = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 1}\), both of \(Q_{2 \rightarrow 3} \text{and}\ Q_{3 \rightarrow 1} = 0\), \(W_{net} = 0\). Therefore, \( \text{Efficiency, } \text{Eff} = \frac{0}{Q_{1 \rightarrow 2}} = 0\).
06

Title - Calculate Carnot Efficiency

Carnot efficiency is given by \(\text{Eff}_{\text{Carnot}} = 1 - \frac{T_{3}}{T_{1}}\). With \(T_{1} = 2000K\), \(T_{3} = T_1 \times \bigg(\frac{P_3}{P_1}\bigg)^{1/4}\rightarrow T_3 = 2000 \times0.616 =1232K\). Thus, \( \text{Eff}_{\text{Carnot}} = 1 - \frac{1232}{2000} = 0.384 = 38.4%\).

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the idealized physical body that absorbs all incident electromagnetic radiation and reflects none. It re-emits this energy in a spectrum that depends solely on the temperature of the body.
The equation of state for blackbody radiation used in heat engines involves key quantities like pressure (P), temperature (T), and volume (V).
In the given problem, the pressure is proportional to the fourth power of temperature, expressed as \(P = \frac{1}{3}aT^{4}\). Consequently, the internal energy (U) of the blackbody radiation is given by \(U = aVT^{4}\).
Here, 'a' is Stefan's constant, whose value is approximately \(7.566 \times 10^{-16} J/(m^{3}K^{4})\).
This relationship is critical in understanding how energy is stored and released within the engine cycle.
  • Blackbody radiations play a crucial part in calculating work done and heat absorbed in various thermodynamic processes.
  • For example, the constant pressure process (1 to 2) shows how volume changes directly relate to temperature changes under constant pressure.
Adiabatic Process
An adiabatic process is a thermodynamic transformation in which no heat is transferred to or from the system.
This implies that any change in the internal energy of the system results solely from work done by or on the system.
For an adiabatic process involving blackbody radiation, the relationship between volumes and temperatures comes from \(TV^{1/3} = \text{constant}\).
In the problem, the compression from process 3 to 1 is adiabatic, implying that the change in internal energy equals the work done by the gas.
Therefore, the work done during this process can be given by \(W_{3 \rightarrow 1} = \text{U}_{\text{3}} - \text{U}_{\text{1}}\).
Where \(U = aVT^{4}\), indicating the adiabatic contraction from volume \(V_{3}\) to \(V_{1}\).
Understanding adiabatic processes also helps in calculating engine efficiencies.
  • The adiabatic process is vital in creating cycles where no heat transfer occurs, affecting the net work done.
  • For instance, the work from contraction contributes exclusively to changes in internal energy and is used to understand the efficiency calculations.
Carnot Engine Efficiency
The Carnot efficiency defines the maximum theoretical efficiency one can achieve in a heat engine operating between two temperatures, T1 (high) and T3 (low).
The formula for Carnot efficiency is \( \text{Eff}_{\text{Carnot}} = 1 - \frac{T_{3}}{T_{1}}\).
Applying this to our problem, with \(T_{1} = 2000K\) and \(T_{3} = 1232K\), we calculate \( \text{Eff}_{\text{Carnot}} = 1 - \frac{1232}{2000} = 0.384\), or 38.4%.
This result represents the upper limit of efficiency any heat engine can achieve under the given temperature limits.
  • This concept is fundamental because it provides an ideal benchmark for comparing real-world engine efficiencies.
  • While actual efficiencies will always be lower due to irreversibilities, the Carnot cycle offers insights into the theoretical limits set by thermodynamics.

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Most popular questions from this chapter

A Carnot engine uses a paramagnetic substance as its working substance. The equation of state is \(M=n D H / T\), where \(M\) is the magnetization, \(H\) is the magnetic field, \(n\) is the number of moles, \(D\) is a constant determined by the type of substance, and \(T\) is the temperature. (a) Show that the internal energy \(U\), and therefore the heat capacity \(C_{M}\), can only depend on the temperature and not the magnetization. Let us assume that \(C_{M}=C=\) constant. (b) Sketch a typical Carnot cycle in the \(M-H\) plane. (c) Compute the total heat absorbed and the total work done by the Carnot engine. (d) Compute the efficiency of the Carnot engine.

For a low-density gas the virial expansion can be terminated at first order in the density and the equation of state is \( P=\frac{N k_{\mathrm{B}} T}{V}\left[1+\frac{N}{V} B_{2}(T)\right] $$ where \)B_{2}(T)\( is the second virial coefficient. The heat capacity will have corrections to its ideal gas value. We can write it in the form $$ C_{V, N}=\frac{3}{2} N k_{\mathrm{B}}-\frac{N^{2} k_{\mathrm{B}}}{V} F(T) $$ (a) Find the form that \)F(T)\( must have in order for the two equations to be thermodynamically consistent. (b) Find \)C_{P, N}$. (c) Find the entropy and internal energy.

A material is found to have a thermal expansivity \(\alpha_{P}=R / P v+a / R T^{2} v\) and an isothermal compressibility \(K_{T}=1 / v(T f(P)+(b / P))\) where \(v=V / n\) is the molar volume. (a) Find \(f(P)\). (b) Find the equation of state. (c) Under what conditions is this material mechanically stable?

Show that \(T \mathrm{~d} s=c_{x}(\partial T / \partial Y)_{x} \mathrm{~d} Y+c_{Y}(\partial T / \partial x)_{Y} \mathrm{~d} x\), where \(x=X / n\) is the amount of extensive variable, \(X\), per mole, \(c_{x}\) is the heat capacity per mole at constant \(x\), and \(c_{Y}\) is the heat capacity per mole at constant \(Y\).

A stochastic process, involving three fluctuating quantities, \(x_{1}, x_{2}\), and \(x_{3}\), has a probability distribution $$ P\left(x_{1}, x_{2}, x_{3}\right)=C \exp \left[-\frac{1}{2}\left(2 x_{1}^{2}+2 x_{1} x_{2}+4 x_{2}^{2}+2 x_{1} x_{3}+2 x_{2} x_{3}+2 x_{3}^{2}\right)\right] $$ where \(C\) is the normalization constant. (a) Write probability distribution in the form \(P\left(x_{1}, x_{2}, x_{3}\right)=C \exp \left(-1 / 2 x^{T} \cdot g+x\right)\), where \(g\) is a \(3 \times 3\) symmetric matrix, \(x\) is a column matrix with matrix elements \(x_{i}, i=1,2,3\), and \(x^{T}\) is its transpose. Obtain the matrix \(\boldsymbol{g}\) and its inverse \(g^{-1}\). (b) Find the eigenvalues \(\lambda_{i}(i=1,2,3)\) and orthonormal eigenvectors of \(\boldsymbol{g}\) and obtain the \(3 \times 3\) orthogonal matrix \(\boldsymbol{O}\) that diagonalizes the matrix \(\boldsymbol{g}\) (get numbers for all of them). Using this orthogonal matrix, we can write \(x^{\mathrm{T}} \cdot g \cdot x=x^{\mathrm{T}} \cdot \boldsymbol{O}^{\mathrm{T}} \cdot \boldsymbol{O} \cdot g \cdot \boldsymbol{O}^{\mathrm{T}} \cdot \boldsymbol{O} \cdot \boldsymbol{x}=\boldsymbol{a}^{\mathrm{T}} \cdot \bar{\Lambda} \cdot \boldsymbol{a}=\sum_{i=1}^{3} \lambda_{i} a_{i}^{2}\) where \(\boldsymbol{O} \cdot g \cdot \boldsymbol{O}^{\mathrm{T}}=\bar{\Lambda}\) is a \(3 \mathrm{x}\) 3 diagonal matrix with matrix elements \((\bar{A})_{i, j}=\lambda_{i} \delta_{i, j}\) and \(\boldsymbol{O} \cdot \boldsymbol{x}=\boldsymbol{a}\) is a column matrix with elements, \(a_{i}(i=1,2,3)\). (c) Compute the normalization constant, C. (d) Compute the moments \(\left(x_{i}\right)(i=1,2,3),\left\langle x_{i} x_{j}\right\rangle(i=1,2,3, j=1,2,3)\left(x_{1}^{2} x_{2} x_{3}\right)\) and \(\left\langle x_{1} x_{2}^{2} x_{3}\right\rangle+\) (Note that Exercises \(\mathrm{A.7}\) and \(\mathrm{A} .8\) might be helpful.)

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