Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S25

For a unit mass of ideal gas, $h=u+P/\rho =u+R_{a}T$. Since $u$ depends only on temperature, $c_p=dh/dT=c_v+R_a$, giving $\boxed{R_a=c_p-c_v}$. For a dry parcel in adiabatic, quasistatic ascent, the first law of thermodynamics is $c_{p}dT={\rho }^{-1}dP$. Environmental hydrostatic equilibrium gives $dP/dz=-\rho g$, so a convectively well-mixed layer has the dry adiabatic lapse rateThis assumes negligible condensation, approximately constant $g$, well-mixed composition and efficient convection; a radiative or stably stratified layer need not follow it. For mole fractions $x_N,x_C$ and molar masses $M_N,M_C$, $R_a=R/(x_N{M}_N+x_C{M}_C)$ and $c_p=[x_N{C}_{p,N}+x_C{C}_{p,C}]/(x_N{M}_N+x_C{M}_C)$. Molecular vibrational excitation can make $c_p$ depend on temperature; then ${\Gamma }_d=g/c_p(T)$. Nitrogen dominance near terrestrial temperatures gives the familiar order $10\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$.

The usual estimate adds a grey, infrared-absorbing radiative equilibrium upper atmosphere to the convective lower atmosphere. Consider a very optically thin upper layer: it receives upwelling infrared flux approximately $F_0$ from below, absorbs a small fraction $d\tau$, and emits $\sigma T^{4}d\tau$ into each hemisphere. Radiative balance givesfor $F_0=300\mathrm{W}{\mathrm{m}}^{-2}$. The corresponding effective emission temperature is $(F_0/\sigma {)}^{1/4}\simeq 270\mathrm{K}$. Without shortwave absorption, the model has no solar-heated stratospheric inversion.

This skin estimate requires the grey absorption/emission and angular approximations just stated. Horizontal mixing and no shortwave absorption alone do not determine an exact minimum for a spectrally selective nitrogen–carbon-dioxide atmosphere: absorption coefficients and spectral windows can change its upper-atmosphere radiative equilibrium.

Downward-from-space optical depth at wavelength $\lambda$ iswhere ${\kappa }_{\lambda }$ is the mass absorption coefficient. A nearly isothermal, hydrostatic equilibrium, compositionally well-mixed atmosphere with height-independent ${\kappa }_{\lambda }$ has $\rho \propto e^{-z/H}$ and $\tau \propto e^{-z/H_{\tau }}$ with $H_{\tau }=H=R_{a}T/g$. Thus temperature, mean molecular mass and gravity set the basic scale height. Changes in the absorber mixing ratio, pressure broadening, temperature dependence of line strengths, collision-induced absorption and departures from hydrostatic equilibrium modify it. For example if local absorption scales as ${\rho }^2$ in an isothermal atmosphere, $H_{\tau }=H/2$.

Use a convective layer $T(z)=T_s-\Gamma z$ capped at an isothermal minimum $T_t$ at height $z_t$. Let the initial optical depth be $\tau (z)={\tau }_s{e}^{-z/H_{\tau }}$ and ${\tau }_t={\tau }_s{e}^{-z_t/H_{\tau }}$. Within the lapse-rate layer, $d{B}_{\lambda }/d\tau =B^{\prime }\Gamma H_{\tau }/\tau$; above it the derivative vanishes. The supplied Schwarzschild equation therefore givesAssume the additional carbon dioxide multiplies the relevant opacity by $a=1.1$, leaving temperatures, the lapse rate and $H_{\tau }$ initially fixed. All material surfaces then acquire $a$ times their previous optical depth. HenceFor ${\tau }_t\ll 1$ and ${\tau }_s\gg 10$, the specified approximation givesFor illustration, $\Gamma =6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ and $H_{\tau }=7\mathrm{km}$ give $-4.34\mathrm{K}$; these are illustrative, not supplied data. The inequalities put the effective emitting level $\tau \sim 1$ inside the decreasing-temperature layer, well above an opaque surface and below the thin isothermal cap. Increasing opacity raises that emitting level by $H_{\tau }\ln 1.1$, so it is colder. In a spectral window or above a temperature inversion the approximation and possibly the sign change. Opacity proportional to mixing ratio is also an assumption, not exact for every absorption mechanism.

$F$ is the imposed radiative forcing; $-\lambda T_s$ is the restoring change in outgoing radiation; and $-\gamma (T_s-T_d)$ transfers heat from the surface layer to the deep ocean, whose equal and opposite gain appears in its own energy balance. The additional $-{\lambda }^{\prime }(T_s-T_d)$ is a radiative adjustment correlated with the surface–deep-ocean temperature contrast, representing a pattern-dependent effective feedback rather than another conserved interlayer heat flux.

For constant forcing, final equilibrium has $T_s=T_d=F/\lambda$. During transient warming $T_d<T_s$, so the radiative response inferred from a recent global energy budget is ${\lambda }_{\mathrm{eff}}=\lambda +{\lambda }^{\prime }(1-T_d/T_s)>\lambda$. If that response were mistaken for the final feedback, the inferred equilibrium climate sensitivity would be too small. Summing the two equations makes the distinction explicit: $d(C_s{T}_s+C_d{T}_d)/dt=F-\lambda T_s-{\lambda }^{\prime }(T_s-T_d)$.

Set ${\lambda }^{\prime }=0$. A decaying normal mode $e^{-kt}$ obeysFor $C_d\gg C_s$, its sum and product of roots giveInitially $T_s=T_d=T_0$, so $a_1+a_2=1$ and $a_1{k}_1+a_2{k}_2=\lambda /C_s$. To leading order,On the slow normal mode, the first equation gives $T_d/T_s=(\lambda +\gamma -C_s{k}_2)/\gamma \simeq (\lambda +\gamma )/\gamma$, verifying proportionality after the fast transient has decayed. The small parameter must also make $k_2/k_1$ small; the supplied positive coefficients and $C_d\gg C_s$ satisfy this.

A linearly increasing radiative forcing produces a fast adjustment over about four years superposed on a slow deep-ocean response over about four hundred years. Once forcing stops increasing at year 70, surface warming continues and approaches the final $3\mathrm{K}$ equilibrium. The right panel of solution gives a model response normalized to the supplied $2\mathrm{K}$ endpoint and $3\mathrm{K}$ equilibrium.

Write $u=t-70$ after stabilization and $3-T_s=A_f{e}^{-u/4}+A_s{e}^{-u/400}$, with $A_f+A_s=1\mathrm{K}$. For a 70-year forcing ramp, the normalized step response is $1-b_f{e}^{-t/4}-b_s{e}^{-t/400}$, $b_f+b_s=1$. The ramp endpoint impliesThis gives $b_s\simeq 0.322$, $A_f\simeq 0.116\mathrm{K}$ and $A_s\simeq 0.884\mathrm{K}$. Thusat $u=10,15,20$ years respectively. The usual order estimate after the fast tail has largely disappeared is $\boxed{{\dot{T}}_s\sim (3-2)/400\simeq 0.0025\mathrm{K}\mathrm{y}{\mathrm{r}}^{-1}}$, with a still visible fast contribution near year 10.

To hold $T_s=T_{\ast }$ fixed, the deep layer obeys $T_s-T_d\propto e^{-\gamma u/C_d}$ and the surface equation requires $F=\lambda T_{\ast }+\gamma (T_{\ast }-T_d)$. Hence $\dot{F}=-{\gamma }^2(T_{\ast }-T_d)/C_d$. Using $b_f\simeq \lambda /(\lambda +\gamma )$ gives $\gamma /\lambda \simeq 0.475$ and $\gamma /C_d\simeq 1/(400b_f)\simeq 0.00369\mathrm{y}{\mathrm{r}}^{-1}$. With $T_{\ast }=2\mathrm{K}$ and initial forcing $F_0=3\lambda$, the initial fractional decline isThe rate declines slowly as the ocean contrast disappears. A coarser estimate treating the post-fast warming as an effective equilibrium change gives $0.0025/3\simeq 0.08\%$ per year; distinguishing the instantaneous surface feedback from the equilibrium feedback explains that difference.

Wording qualification: the printed text says a “1% increase … over 70 years”. The conventional transient climate response experiment uses $1\%$ per year, which nearly doubles concentration in 70 years. A literal total $1\%$ increase would not define the usual equilibrium climate sensitivity or transient climate response. The supplied $3\mathrm{K}$ and $2\mathrm{K}$ are used here as the endpoint normalizations, and a logarithmically uniform concentration growth gives the assumed linear forcing ramp.

The transient climate response to cumulative carbon emissions is $\boxed{\kappa \simeq \Delta T/\mathcal{E}}$, where $\mathcal{E}$ is cumulative emitted carbon dioxide in a stated mass unit. Here the unit is tonnes of $\mathrm{C}{\mathrm{O}}_2$, rather than tonnes of carbon. Increasing concentration weakens the incremental radiative forcing per added atmospheric tonne because forcing is approximately logarithmic. Over a substantial range this is partly offset by a weakening of carbon sinks and by the changing ratio between transient warming and forcing as ocean heat uptake evolves. These approximate compensations make warming nearly proportional to cumulative emissions; no exact cancellation follows from the logarithmic forcing law alone. The approximation is range- and pathway-dependent, and is not a theorem valid for arbitrary emissions histories.

An additional tonne changes eventual temperature by approximately $\delta T=\kappa$. Linearizing the damage function gives $\delta L(t^{\prime })=\kappa \gamma D_{0}W(t^{\prime })T(t^{\prime }{)}^{\gamma -1}$. Put $s=t^{\prime }-t$, $W(t+s)=W(t)e^{gs}$, and hold $T$ approximately fixed across the effective discount horizon. The discounted integral issoIf $r\le g$, this infinite-horizon approximation diverges; a finite result then needs different growth, damages or discounting assumptions. It also treats the marginal warming from the extra tonne as effectively persistent over the relevant horizon.

Equating marginal mitigation cost to the previous marginal damage givesUsing $\dot{T}=\kappa E$ from the cumulative-emissions approximation,Thus $T\propto t^{1/\gamma }$ at late times and $E\propto t^{-(\gamma -1)/\gamma }$. Warming slows but never stops, and cumulative emissions remain unbounded. This follows from a cost model in which zero emissions is infinitely costly and from a marginal-damage rule with constant positive $r-g$.

The approximation eventually fails if cumulative emissions leave the range of constant TCRE, if large damages suppress economic growth, if technological change alters $M_0$, or if fossil resources and mitigation options impose different limits. The slowly varying-temperature approximation also needs $\dot{T}/[T(r-g)]\ll 1$. Thresholds, adaptation, uncertainty and unequal impacts are omitted; the formula is a property of this model rather than a universal policy trajectory.

With $\gamma =2$ and $T(t+s)=T(t)+\dot{T}s$, the marginal damage is $2\kappa D_{0}W(t)e^{gs}[T+\dot{T}s]$. Integrating the constant and linear terms givesThe prefactor is $2(0.5\times 10^{-12})(0.003)(90\times 10^{12})=0.27$ in the corresponding dollar–temperature–time units. For the historical scenario with $r=0.03$ and $g=0.02$, the bracket is $1/0.01+0.02/0.01^2=300$, so $\boxed{S=81\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$. For $r=0.07$, $r-g=0.05$, giving global $S=0.27(20+8)=7.56$ dollars per tonne. Restricting impacts to ten per cent gives $\boxed{S_{\mathrm{domestic}}=0.756\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$.

The factor of about 107 reduction comes from both the higher discount rate and the geographical restriction; it is not a change in climate physics. The administration descriptions are those of the 2019 question, not claims about present policy. With a net discount rate of only one per cent, a century-scale horizon contributes substantially, making the linear warming and exponential income extrapolations consequential assumptions.