Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S25 / 2 / 3 / Solution

At $z=4\mathrm{km}$, the snow gain is $1-0.2(4)=0.2$ and the model's top Celsius temperature is $275-273.15-6(4)=-22.15^{\circ }\mathrm{C}$. Equilibrium givesin the normalized rate convention. Using $273\mathrm{K}$ for the Celsius zero instead would give $4.4^{\circ }\mathrm{C}$. The local slope is $-0.2+4.43(6)/(22.15{)}^2\simeq -0.1458\mathrm{k}{\mathrm{m}}^{-1}<0$, as required. Initially $z_{-}=1.3083\mathrm{km}$ and $z_{\mathrm{slope}}\simeq 2.230\mathrm{km}$, while $z_{+}=4\mathrm{km}$.

As $T_s$ slowly rises, melt loss increases at fixed $z$, the stable high branch moves downward and the unstable branch upward. They merge atBeyond the fold the model has no steady cold ice cap: its net balance is negative throughout the physical interval and ice shrinks until the assumed law ceases to apply. Quasistatic tracking presumes adjustment is fast compared with the imposed warming, an assumption that deteriorates near the fold as the restoring slope tends to zero. This singular idealized law is not a quantitative real-world melting forecast.