IPhO 2024 · Theory problem 1

The Greenhouse Effect

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In 2021, Syukuro Manabe and Klaus Hasselmann were awarded half of the Nobel Prize in Physics for their work in modeling Earth's climate and accurately predicting the global warming caused by human industrial activities. In this problem, we will examine a simple model of global warming due to the greenhouse effect. The greenhouse gases alter the optical properties of the Earth's atmosphere in transmitting or absorbing Earth's infrared radiation, resulting in a rise in the average temperature of the planet.

All objects, at different temperatures, emit thermal radiation. The quantity \(u(\lambda,T)d\lambda\) indicates the thermal radiative power per unit area of an object at temperature \(T\) between the wavelengths \(\lambda\) and \(\lambda+d\lambda\). According to Planck’s theory of blackbody radiation, we have

\[ u(\lambda,T)=\frac{2\pi hc^2}{\lambda^5} \frac{1}{\exp\!\left(\frac{hc}{\lambda k_{\mathrm B}T}\right)-1}, \tag{1} \]

in which \(hc=1.24\times10^3\ \mathrm{eV\cdot nm}\) and \(k_{\mathrm B}=8.62\times10^{-5}\ \mathrm{eV/K}\). The wavelength corresponding to the maximum of \(u(\lambda,T)\) comes from the relation \(\lambda_{\max}T=b\) (Wien’s displacement law). Indeed, using equation (1), it can be shown that \(b=\frac{hc}{x_mk_{\mathrm B}}\), where the dimensionless quantity \(x_m\) is the non-trivial root of an equation of the form \(f(x)=0\); you are asked to find the function \(f(x)\) in one of the following tasks. Total radiative power per unit area of a blackbody in all wavelengths is given by the Stephan-Boltzmann law as \(U(T)=\sigma T^4\) where \(\sigma=5.67\times10^{-8}\ \mathrm{W/m^2K^4}\). Moreover, according to Kirchhoff’s law of radiation, at thermal equilibrium a body absorbing a certain fraction of the incident radiation at a specific wavelength, will radiate the same fraction of the blackbody radiation at that same wavelength.

Throughout this problem assume that the Sun is a blackbody at its average surface temperature of \(T_{\mathrm S}=5.77\times10^3\ \mathrm K\). The Sun’s radius is \(R_{\mathrm S}=6.96\times10^8\ \mathrm m\) and the average distance between the Earth and the Sun is \(d=1.50\times10^{11}\ \mathrm m\). We denote by \(\tilde{u}_{\mathrm S}(\lambda)\), the spectral solar power radiated into a unit area of the Earth normal to the direction of radiation. The integral of this quantity over all wavelengths, i.e. \(S_0=\int\tilde{u}_{\mathrm S}(\lambda)d\lambda\), is called the solar constant.

In this problem assume that the Earth is in thermal equilibrium and has the same temperature at all points on its surface. In all parts of the problem, express the desired quantity in parametric form in terms of the data given in the problem and then find its numerical value accurate to three significant figures. The required units are indicated on the answer sheet.

A. Earth as a Blackbody

In this part, consider the Earth’s surface as a blackbody and neglect the Earth's atmosphere.

Part Task Points
A-1 Find the solar constant, \(S_0\). 0.6 pt
A-2 Find the Earth’s temperature, \(T_{\mathrm E}\). 0.6 pt
A-3 Find the function \(f(x)\). 0.4 pt
A-4 Calculate the numerical value of \(x_m\), and from this value \(x_m\), find the value of \(b\). 0.4 pt
A-5 Find \(\lambda_{\max}\) for the Sun and the Earth. 0.2 pt

In figure 1 the functions \(\gamma\tilde{u}_{\mathrm S}(\lambda)\) and \(u(\lambda,T_{\mathrm E})\) are plotted versus \(\lambda\), where \(\gamma\) is a dimensionless coefficient to rescale \(\tilde{u}_{\mathrm S}(\lambda)\) such that the values of the two peaks coincide.

Part Task Points
A-6 Determine \(\gamma\). 0.8 pt
Figure 1

Figure 1 - The plot of \(u(\lambda,T_{\mathrm E})\) (red) and \(\gamma\tilde{u}_{\mathrm S}(\lambda)\) (blue) versus \(\lambda\)

B. The Greenhouse Effect

In this part, we introduce a simple model in which the Earth’s atmosphere is modeled as a thin layer at a small distance above the Earth’s surface so that the difference between the area of the atmosphere’s layer and the area of the Earth's surface can be neglected (see figure 2). In what follows assume that the major part of the thermal radiation from the Earth and the Sun are emitted at wavelengths near the \(\lambda_{\max}\) for each one. Also assume that the “atmosphere layer” reflects a fraction \(r_{\mathrm A}=0.255\) of the visible-ultraviolet radiation incident from above or below, and completely transmits the rest. Assume that the atmosphere does not reflect any part of the infrared radiation, however, it absorbs a fraction \(\varepsilon\) of the infrared radiation and transmits the rest. This behavior, known as the greenhouse effect, changes the average temperature of the Earth. The Earth’s surface, on the other hand, reflects a fraction \(r_{\mathrm E}\) of the visible-ultraviolet radiation and absorbs the rest of this radiation and all the infrared radiation.

Figure 2

Figure 2 - Thermal flows between the Earth and the atmosphere

Part Task Points
B-1 Assume that \(\varepsilon=1\) and \(r_{\mathrm E}=0\), and calculate the Earth’s temperature \(T_{\mathrm E}\) and the atmosphere’s temperature \(T_{\mathrm A}\). 1.0 pt

Now assume that \(r_{\mathrm E}\ne0\). In this case, the combined system of “Earth + atmosphere” reflects a different fraction of the solar radiation, called “albedo” and denoted by \(\alpha\).

Part Task Points
B-2 Determine the albedo, \(\alpha\), in terms of \(r_{\mathrm E}\) and \(r_{\mathrm A}\). Then calculate its numerical value assuming \(r_{\mathrm E}=0.102\) (and \(r_{\mathrm A}=0.255\)). 1.6 pt
B-3 a) Express the Earth’s temperature in terms of \(\sigma\), \(\alpha\), \(S_0\), and \(\varepsilon\).

b) Using the given data and the calculated albedo, find the numerical value of \(\varepsilon\) which leads to the current average temperature of \(T_{\mathrm E}=288\ \mathrm K\) for the Earth.		 1.0 pt
B-4 Find \(\frac{dT_{\mathrm E}}{d\varepsilon}\) and determine by how much the Earth’s temperature increases if \(\varepsilon\) increases by one percent. 0.8 pt

Assume \(T_{\mathrm A}=245\ \mathrm K\) and \(T_{\mathrm E}=288\ \mathrm K\). These values come from real data and may differ from the results which you have obtained in the previous tasks. Now suppose that a non-radiative (e. g. convective) thermal flow \(J_{\mathrm{NR}}=k(T_{\mathrm E}-T_{\mathrm A})\) is maintained from the Earth to the atmosphere, where \(k\) is a constant. The quantity, \(J_{\mathrm{NR}}\), is the transmitted power per unit area.

Part Task Points
B-5 Calculate \(\varepsilon\) and \(k\) in terms of \(T_{\mathrm E}\), \(T_{\mathrm A}\), \(\sigma\), \(\alpha\), and \(S_0\). 1.6 pt
B-6 a) Differentiating the equations obtained in part B-5 with respect to \(\varepsilon\), find the two algebraic equations satisfied by \(\frac{dT_{\mathrm A}}{d\varepsilon}\) and \(\frac{dT_{\mathrm E}}{d\varepsilon}\).

b) Use these equations to find the numerical value of change in the Earth’s temperature as a result of a one percent increase in the value of \(\varepsilon\).		 1.0 pt

Official English text of the problem.

Answers

Define the globally averaged absorbed solar flux by \[ Q=\frac{(1-\alpha)S_0}{4}. \] A one-percent increase is taken to mean \(\Delta\varepsilon=0.01\varepsilon\). Numerical results use unrounded intermediate values.

  • A-1 \[ \boxed{S_0=\sigma T_{\mathrm S}^4\left(\frac{R_{\mathrm S}}d\right)^2 =1.35\times10^3\ \mathrm{W\,m^{-2}}.} \]

  • A-2 \[ \boxed{T_{\mathrm E}=\left(\frac{S_0}{4\sigma}\right)^{1/4} =T_{\mathrm S}\sqrt{\frac{R_{\mathrm S}}{2d}} =278\ \mathrm K.} \]

  • A-3 \[ \boxed{f(x)=5(1-e^{-x})-x.} \]

  • A-4 \[ \boxed{x_m=4.97,\qquad b=\frac{hc}{x_mk_{\mathrm B}}=2.90\times10^{-3}\ \mathrm{m\,K}.} \]

  • A-5 \[ \boxed{\lambda_{\max,\mathrm S}=\frac b{T_{\mathrm S}}=502\ \mathrm{nm}, \qquad \lambda_{\max,\mathrm E}=\frac b{T_{\mathrm E}}=10.4\ \mu\mathrm m.} \]

  • A-6 \[ \boxed{\gamma=\left(\frac d{R_{\mathrm S}}\right)^2 \left(\frac{T_{\mathrm E}}{T_{\mathrm S}}\right)^5 =\frac14\sqrt{\frac{R_{\mathrm S}}{2d}} =1.20\times10^{-2}.} \]

  • B-1 \[ \boxed{T_{\mathrm E}= \left[\frac{(1-r_{\mathrm A})S_0}{2\sigma}\right]^{1/4} =307\ \mathrm K,} \] \[ \boxed{T_{\mathrm A}= \left[\frac{(1-r_{\mathrm A})S_0}{4\sigma}\right]^{1/4} =258\ \mathrm K.} \]

  • B-2 \[ \boxed{\alpha=r_{\mathrm A}+ \frac{r_{\mathrm E}(1-r_{\mathrm A})^2}{1-r_{\mathrm A}r_{\mathrm E}} =0.313.} \]

  • B-3(a) \[ \boxed{T_{\mathrm E}= \left[\frac{(1-\alpha)S_0}{4\sigma(1-\varepsilon/2)}\right]^{1/4}.} \]

  • B-3(b) \[ \boxed{\varepsilon= 2-\frac{(1-\alpha)S_0}{2\sigma T_{\mathrm E}^4} =0.809.} \]

  • B-4 \[ \boxed{\frac{dT_{\mathrm E}}{d\varepsilon} =\frac{T_{\mathrm E}}{4(2-\varepsilon)} =60.4\ \mathrm K,} \qquad \boxed{\Delta T_{\mathrm E}\simeq0.489\ \mathrm K.} \] The temperature change is the first-order result; the exact finite change is \(0.491\ \mathrm K\).

  • B-5 \[ \boxed{\varepsilon= \frac{\sigma T_{\mathrm E}^4-Q} {\sigma(T_{\mathrm E}^4-T_{\mathrm A}^4)} =0.849,} \] \[ \boxed{k= \frac{(\sigma T_{\mathrm E}^4-Q)(2T_{\mathrm A}^4-T_{\mathrm E}^4)} {(T_{\mathrm E}^4-T_{\mathrm A}^4)(T_{\mathrm E}-T_{\mathrm A})} =0.365\ \mathrm{W\,m^{-2}\,K^{-1}}.} \]

  • B-6(a) With \(X=dT_{\mathrm E}/d\varepsilon\) and \(Y=dT_{\mathrm A}/d\varepsilon\), \[ \boxed{(4\sigma T_{\mathrm E}^3+k)X -(4\varepsilon\sigma T_{\mathrm A}^3+k)Y =\sigma T_{\mathrm A}^4,} \] \[ \boxed{-(4\varepsilon\sigma T_{\mathrm E}^3+k)X +(8\varepsilon\sigma T_{\mathrm A}^3+k)Y =\sigma(T_{\mathrm E}^4-2T_{\mathrm A}^4).} \]

  • B-6(b) \[ \boxed{X=61.7\ \mathrm K,\qquad Y=47.8\ \mathrm K,\qquad \Delta T_{\mathrm E}\simeq0.524\ \mathrm K.} \] The temperature change is the first-order result; the exact finite change is \(0.527\ \mathrm K\).

Solution

A-1

The Sun’s total radiated power follows from the Stefan–Boltzmann law: \[ L_{\mathrm S}=4\pi R_{\mathrm S}^2\sigma T_{\mathrm S}^4. \] Conservation of radiative power distributes this luminosity over a sphere of radius \(d\). Consequently, \[ S_0=\frac{L_{\mathrm S}}{4\pi d^2} =\sigma T_{\mathrm S}^4\left(\frac{R_{\mathrm S}}d\right)^2. \] Substitution gives \[ S_0=1353.076\ldots\ \mathrm{W\,m^{-2}} =\boxed{1.35\times10^3\ \mathrm{W\,m^{-2}}}. \]

The same geometrical dilution applies at every wavelength: \[ \tilde{u}_{\mathrm S}(\lambda) =\left(\frac{R_{\mathrm S}}d\right)^2u(\lambda,T_{\mathrm S}). \]

A-2

Let \(R_{\mathrm E}\) denote the Earth’s radius. The Earth intercepts sunlight over its projected area \(\pi R_{\mathrm E}^2\), while its entire surface \(4\pi R_{\mathrm E}^2\) emits thermal radiation.

For a blackbody Earth in equilibrium, \[ \pi R_{\mathrm E}^2S_0 =4\pi R_{\mathrm E}^2\sigma T_{\mathrm E}^4. \] Thus \[ T_{\mathrm E} =\left(\frac{S_0}{4\sigma}\right)^{1/4} =T_{\mathrm S}\sqrt{\frac{R_{\mathrm S}}{2d}} =277.920\ldots\ \mathrm K. \] Therefore, \[ \boxed{T_{\mathrm E}=278\ \mathrm K}. \]

A-3

At fixed \(T\), introduce \[ x=\frac{hc}{\lambda k_{\mathrm B}T}. \] Logarithmic differentiation of Planck’s law gives \[ \frac{1}{u}\frac{\partial u}{\partial\lambda} =\frac1\lambda\left[-5+\frac{x e^x}{e^x-1}\right]. \] At a finite-wavelength maximum this derivative vanishes, so \[ 5=\frac{x}{1-e^{-x}}. \] Hence a suitable function is \[ \boxed{f(x)=5(1-e^{-x})-x}. \]

The algebraic root \(x=0\) corresponds to \(\lambda\to\infty\), rather than a finite-wavelength maximum. The required root is positive and nonzero.

A-4

The nonzero root can be obtained, for example, using Newton’s method: \[ x_{n+1} =x_n-\frac{5(1-e^{-x_n})-x_n}{5e^{-x_n}-1}. \] Starting at \(x_0=5\) gives \[ x_m=4.965114\ldots. \] Since \(x_m=hc/(\lambda_{\max}k_{\mathrm B}T)\), \[ \lambda_{\max}T=\frac{hc}{x_mk_{\mathrm B}}=b. \] Using the supplied constants, \[ b= \frac{1.24\times10^3} {(4.965114\ldots)(8.62\times10^{-5})} \ \mathrm{nm\,K} =2.897244\ldots\times10^{-3}\ \mathrm{m\,K}. \] Therefore, \[ \boxed{x_m=4.97,\qquad b=2.90\times10^{-3}\ \mathrm{m\,K}}. \]

Planck’s function tends to zero at both wavelength endpoints, and this is its unique finite positive stationary point, so it is the maximum.

A-5

Applying Wien’s displacement law to each body, \[ \lambda_{\max,\mathrm S} =\frac b{T_{\mathrm S}} =502.122\ldots\ \mathrm{nm}, \] \[ \lambda_{\max,\mathrm E} =\frac b{T_{\mathrm E}} =10.4247\ldots\ \mu\mathrm m. \] Thus \[ \boxed{\lambda_{\max,\mathrm S}=502\ \mathrm{nm},\qquad \lambda_{\max,\mathrm E}=10.4\ \mu\mathrm m}. \]

The solar peak is in the visible range, while the Earth’s peak is in the infrared.

A-6

At the maximum, \(\lambda_{\max}=b/T\) and the exponential argument equals \(x_m\). Therefore, \[ u_{\max}(T) =\frac{2\pi hc^2}{b^5(e^{x_m}-1)}T^5. \] Thus the peak spectral power per unit wavelength scales as \(T^5\).

Equality of the two plotted peak heights requires \[ \gamma\left(\frac{R_{\mathrm S}}d\right)^2u_{\max}(T_{\mathrm S}) =u_{\max}(T_{\mathrm E}), \] which gives \[ \gamma=\left(\frac d{R_{\mathrm S}}\right)^2 \left(\frac{T_{\mathrm E}}{T_{\mathrm S}}\right)^5. \] From A-2, \[ \left(\frac{T_{\mathrm E}}{T_{\mathrm S}}\right)^4 =\frac14\left(\frac{R_{\mathrm S}}d\right)^2. \] Consequently, \[ \gamma=\frac{T_{\mathrm E}}{4T_{\mathrm S}} =\frac14\sqrt{\frac{R_{\mathrm S}}{2d}} =\boxed{1.20\times10^{-2}}. \]

B-1

In part B, the stated spectral approximation treats solar radiation as visible-ultraviolet and thermal emission from the Earth and atmosphere as infrared. The atmosphere absorbs no solar radiation. By Kirchhoff’s law, its infrared emissivity equals \(\varepsilon\), so each face emits a flux \(\varepsilon\sigma T_{\mathrm A}^4\). The Earth’s infrared emissivity is unity.

Here \(\varepsilon=1\) and \(r_{\mathrm E}=0\). The globally averaged solar flux absorbed at the surface is \[ Q_1=\frac{(1-r_{\mathrm A})S_0}{4}. \] Atmospheric equilibrium requires \[ \sigma T_{\mathrm E}^4=2\sigma T_{\mathrm A}^4, \] because the atmosphere absorbs the Earth’s upward emission and emits in both directions.

The surface balance is \[ Q_1+\sigma T_{\mathrm A}^4=\sigma T_{\mathrm E}^4. \] Combining the equations, \[ \sigma T_{\mathrm A}^4=Q_1,\qquad \sigma T_{\mathrm E}^4=2Q_1. \] Therefore, \[ T_{\mathrm A} =\left[\frac{(1-r_{\mathrm A})S_0}{4\sigma}\right]^{1/4} =\boxed{258\ \mathrm K}, \] \[ T_{\mathrm E} =\left[\frac{(1-r_{\mathrm A})S_0}{2\sigma}\right]^{1/4} =\boxed{307\ \mathrm K}. \]

B-2

Normalize the incoming solar power to unity. The atmosphere immediately reflects a fraction \(r_{\mathrm A}\). The fraction escaping to space after one reflection from the surface is \[ (1-r_{\mathrm A})r_{\mathrm E}(1-r_{\mathrm A}). \] Each additional round trip between the surface and atmosphere multiplies the remaining reflected power by \(r_{\mathrm A}r_{\mathrm E}\).

Summing all contributions, \[ \begin{aligned} \alpha &=r_{\mathrm A} +r_{\mathrm E}(1-r_{\mathrm A})^2 \sum_{n=0}^{\infty}(r_{\mathrm A}r_{\mathrm E})^n\\ &=r_{\mathrm A} +\frac{r_{\mathrm E}(1-r_{\mathrm A})^2} {1-r_{\mathrm A}r_{\mathrm E}}. \end{aligned} \] Equivalently, \[ 1-\alpha =\frac{(1-r_{\mathrm A})(1-r_{\mathrm E})} {1-r_{\mathrm A}r_{\mathrm E}}. \] With the given reflectivities, \[ \alpha=0.313124\ldots, \qquad \boxed{\alpha=0.313}. \]

All solar power that is not reflected is absorbed by the surface.

B-3

(a) Define \[ Q=\frac{(1-\alpha)S_0}{4}. \] Without non-radiative heat transport, the surface and atmospheric balances are \[ Q+\varepsilon\sigma T_{\mathrm A}^4=\sigma T_{\mathrm E}^4, \] \[ \varepsilon\sigma T_{\mathrm E}^4 =2\varepsilon\sigma T_{\mathrm A}^4. \] For \(\varepsilon>0\), the second equation gives \[ T_{\mathrm A}^4=\frac{T_{\mathrm E}^4}{2}. \] Substituting into the first equation, \[ Q=\sigma T_{\mathrm E}^4\left(1-\frac{\varepsilon}{2}\right). \] Hence \[ \boxed{ T_{\mathrm E} =\left[\frac{(1-\alpha)S_0} {4\sigma(1-\varepsilon/2)}\right]^{1/4}. } \] This expression also extends continuously to \(\varepsilon=0\).

(b) Rearranging for the absorptivity, \[ \varepsilon =2\left(1-\frac{Q}{\sigma T_{\mathrm E}^4}\right) =2-\frac{(1-\alpha)S_0}{2\sigma T_{\mathrm E}^4}. \] The unrounded fluxes are \[ Q=232.348778\ldots\ \mathrm{W\,m^{-2}}, \qquad \sigma(288\ \mathrm K)^4=390.079395\ldots\ \mathrm{W\,m^{-2}}. \] Thus \[ \varepsilon=0.808710\ldots, \qquad \boxed{\varepsilon=0.809}. \]

B-4

Hold \(S_0\) and \(\alpha\) fixed. Logarithmic differentiation of the B-3 result gives \[ \frac{1}{T_{\mathrm E}}\frac{dT_{\mathrm E}}{d\varepsilon} =\frac{1}{8(1-\varepsilon/2)} =\frac{1}{4(2-\varepsilon)}. \] Therefore, \[ \boxed{\frac{dT_{\mathrm E}}{d\varepsilon} =\frac{T_{\mathrm E}}{4(2-\varepsilon)}}. \] At the B-3 equilibrium, \[ \frac{dT_{\mathrm E}}{d\varepsilon} =60.4387\ldots\ \mathrm K =\boxed{60.4\ \mathrm K}. \]

For a relative one-percent increase, \[ \Delta\varepsilon=0.01\varepsilon. \] To first order, \[ \Delta T_{\mathrm E} \simeq\frac{dT_{\mathrm E}}{d\varepsilon}\Delta\varepsilon =\frac{0.01\varepsilon T_{\mathrm E}}{4(2-\varepsilon)} =\boxed{0.489\ \mathrm K}. \]

For comparison, evaluating the finite change directly gives \[ \Delta T_{\mathrm E}^{\mathrm{finite}} =T_{\mathrm E} \left[ \left(\frac{2-\varepsilon}{2-1.01\varepsilon}\right)^{1/4}-1 \right] =\boxed{0.491\ \mathrm K}. \]

B-5

Include the upward non-radiative flux \[ J_{\mathrm{NR}}=k(T_{\mathrm E}-T_{\mathrm A}). \] The surface loses this energy, while the atmosphere gains it. Their equilibrium equations become \[ Q+\varepsilon\sigma T_{\mathrm A}^4 =\sigma T_{\mathrm E}^4+k(T_{\mathrm E}-T_{\mathrm A}), \tag{2} \] \[ \varepsilon\sigma T_{\mathrm E}^4+k(T_{\mathrm E}-T_{\mathrm A}) =2\varepsilon\sigma T_{\mathrm A}^4. \tag{3} \]

For brevity, write \[ E=\sigma T_{\mathrm E}^4,\qquad A=\sigma T_{\mathrm A}^4. \] Adding the two energy balances eliminates the internal heat transfer: \[ Q=(1-\varepsilon)E+\varepsilon A. \] The right-hand side is the radiation escaping to space: transmitted surface radiation plus upward atmospheric emission. Solving, \[ \boxed{\varepsilon=\frac{E-Q}{E-A} =\frac{\sigma T_{\mathrm E}^4-Q} {\sigma(T_{\mathrm E}^4-T_{\mathrm A}^4)}}. \]

Equation (3) gives \[ k(T_{\mathrm E}-T_{\mathrm A})=\varepsilon(2A-E). \] Therefore, \[ \boxed{ k=\frac{\varepsilon\sigma(2T_{\mathrm A}^4-T_{\mathrm E}^4)} {T_{\mathrm E}-T_{\mathrm A}} =\frac{(\sigma T_{\mathrm E}^4-Q)(2T_{\mathrm A}^4-T_{\mathrm E}^4)} {(T_{\mathrm E}^4-T_{\mathrm A}^4)(T_{\mathrm E}-T_{\mathrm A})}. } \]

At \(T_{\mathrm E}=288\ \mathrm K\) and \(T_{\mathrm A}=245\ \mathrm K\), \[ E=390.079395\ldots\ \mathrm{W\,m^{-2}}, \qquad A=204.290135\ldots\ \mathrm{W\,m^{-2}}. \] Consequently, \[ \boxed{\varepsilon=0.849,\qquad k=0.365\ \mathrm{W\,m^{-2}\,K^{-1}}}. \] The resulting heat flux is \[ J_{\mathrm{NR}}=15.7\ \mathrm{W\,m^{-2}}, \] which is positive in the stipulated upward direction.

B-6

(a) When \(\varepsilon\) varies, hold \(S_0\), \(\alpha\), and \(k\) fixed, and allow both temperatures to change. Set \[ X=\frac{dT_{\mathrm E}}{d\varepsilon}, \qquad Y=\frac{dT_{\mathrm A}}{d\varepsilon}. \]

Differentiating equation (2) gives \[ \sigma T_{\mathrm A}^4 +4\varepsilon\sigma T_{\mathrm A}^3Y =4\sigma T_{\mathrm E}^3X+k(X-Y). \] Hence \[ \boxed{ (4\sigma T_{\mathrm E}^3+k)X -(4\varepsilon\sigma T_{\mathrm A}^3+k)Y =\sigma T_{\mathrm A}^4. } \]

Differentiating equation (3) gives \[ \sigma T_{\mathrm E}^4 +4\varepsilon\sigma T_{\mathrm E}^3X+k(X-Y) =2\sigma T_{\mathrm A}^4 +8\varepsilon\sigma T_{\mathrm A}^3Y. \] Thus \[ \boxed{ -(4\varepsilon\sigma T_{\mathrm E}^3+k)X +(8\varepsilon\sigma T_{\mathrm A}^3+k)Y =\sigma(T_{\mathrm E}^4-2T_{\mathrm A}^4). } \]

(b) To give an explicit parametric expression, define \[ p=4\sigma T_{\mathrm E}^3,\qquad q=4\sigma T_{\mathrm A}^3, \] and retain \(E=\sigma T_{\mathrm E}^4\), \(A=\sigma T_{\mathrm A}^4\). The determinant of the system is \[ D=(p+k)(2\varepsilon q+k)-(\varepsilon q+k)(\varepsilon p+k) =\varepsilon(2-\varepsilon)pq +k\bigl[(1-\varepsilon)p+\varepsilon q\bigr]. \] Eliminating \(Y\) gives \[ \boxed{ X=\frac{\varepsilon qE+k(E-A)}{D}. } \] Similarly, \[ Y=\frac{(p+k)(E-2A)+(\varepsilon p+k)A}{D}. \]

Using the equilibrium values from B-5, \[ X=61.7352\ldots\ \mathrm K, \qquad Y=47.7734\ldots\ \mathrm K. \] For \(\Delta\varepsilon=0.01\varepsilon\), \[ \Delta T_{\mathrm E} \simeq X\Delta\varepsilon =(61.7352\ldots)(0.01)(0.84897597\ldots)\ \mathrm K. \] Therefore, \[ \boxed{\Delta T_{\mathrm E}\simeq0.524\ \mathrm K}. \]

Solving the original nonlinear balances at \(\varepsilon_{\mathrm{new}}=1.01\varepsilon\), with the same \(k\), gives the finite change \[ \boxed{\Delta T_{\mathrm E}^{\mathrm{finite}}=0.527\ \mathrm K}. \]

The solution was checked against the official solution.