Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S25 / 2

For dry ascent, $c_{p}dT=Vdp$ and hydrostatic pressure change is $dp/dz=-\rho g=-g/V$, so $\boxed{{\Gamma }_d=-dT/dz=g/c_p}$. During condensation $dq<0$, and $-Ldq>0$ offsets expansion cooling, making the saturated ascent less steep.

For a saturated parcel with $q=q_s(T,p)$, use $dq=q_{T}dT+q_{p}dp$ to obtainWith dilute vapor, $q_p\simeq -q_s/p$ and $q_T\simeq L{q}_s/(R_v{T}^2)$, this becomes ${\Gamma }_d[1+L{q}_s/(R_{a}T)]/[1+L^2{q}_s/(c_p{R}_v{T}^2)]$. In terrestrial conditions the denominator's latent-heating correction dominates. Warming increases saturation humidity strongly, so latent heat increasingly compensates expansion cooling and the moist lapse rate falls. This is a statement about saturated convective adjustment, not every instantaneous environmental temperature profile.

Effective emission temperature satisfies $\sigma T_e^4=S_0(1-\alpha )/4$: absorbed sunlight over the planetary disc balances radiation over the whole sphere. Thus $T_e=[S_0(1-\alpha )/(4\sigma ){]}^{1/4}$. Taking $S_0=1360$ W m${}^{-2}$ gives about 254.5 K. If only a uniform atmospheric/surface temperature shift is allowed, the blackbody response coefficient is ${\lambda }_P=4\sigma T_e^3\simeq 3.74$ W m${}^{-2}$ K${}^{-1}$. HenceThe surface and troposphere shift together under this fixed-profile approximation; their absolute temperatures need not equal $T_e$. A more detailed spectral Planck feedback requires additional information not supplied by the simple model.

At a fixed surface warming, a smaller lapse rate makes the upper troposphere warmer than in an unchanged-lapse profile. Its emission to space therefore increases, giving a negative feedback. Water vapor has the opposite effect under the given coefficients. The net restoring coefficient is ${\lambda }_P-2+0.5\simeq 2.24$ W m${}^{-2}$ K${}^{-1}$, so $\boxed{\Delta T\simeq 1.65\mathrm{K}}$. This calculation uses the same $S_0$ and linearized Planck model as the preceding part.

Albedo reduction increases absorbed short-wave flux by $S_0(a_{1}T+a_2{T}^2)/4$. Define $A={\lambda }_P-1.5-S_0{a}_1/4$ and $B=S_0{a}_2/4$. The perturbation energy balance is $N(T)=F-AT+B{T}^2$, where $F=3.7$ W m${}^{-2}$ for doubling. With the preceding constants, $A\simeq 1.220$ W m${}^{-2}$ K${}^{-1}$ and $B=0.068$ W m${}^{-2}$ K${}^{-2}$. The two roots areStability follows from $N^{\prime }(T)=-A+2BT<0$ at the lower root. They merge at $T=A/(2B)\simeq 8.97$ K and $F_{\max }=A^2/(4B)\simeq 5.47$ W m${}^{-2}$; at equality the equilibrium is marginal, not linearly stable.

Assuming the usual logarithmic forcing extension $F=3.7\log (C/280)/\log 2$ gives a stable branch only for $\boxed{C<C_{\mathrm{crit}}\simeq 7.8\times 10^2\mathrm{ppm}}$. This “maximum” is the supremum of stable concentrations in the stated quadratic, linearized model. It is not a physical prediction of a real-world runaway threshold; extrapolating its feedback laws far beyond their small-change regime is an additional assumption.