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 / 3

Complete oxidation of $10^{12}\mathrm{kg}$ of carbon adds $n_C=10^{12}/(12\times 10^{-3})=8.33\times 10^{13}\mathrm{mol}$ of carbon dioxide. With atmospheric mass about $5.1\times 10^{18}\mathrm{kg}$ and mean air molar mass $28.97\times 10^{-3}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, there are about $1.76\times 10^{20}\mathrm{mol}$ of air. The dry-air mole fraction, approximately the volume fraction, changes byIt is the carbon inventory that counts: oxidation adds oxygen to each molecule but does not change the number of carbon-containing molecules. Using $5.15\times 10^{18}\mathrm{kg}$ instead changes the estimate by only about one percent. The atmospheric constants are standard Earth estimates, not additional data read from a missing plot.^[ref]

The term $b_{1}E$ is the effective atmospheric addition from ongoing emissions. Its coefficient $0.25ppmv/GtC$ is about half the full-retention estimate $0.47ppmv/GtC$, so the model implicitly removes or partitions some emitted carbon promptly. The term $-b_0(C-C_0)$ represents slower net uptake of the excess by other reservoirs; its relaxation time is $b_0^{-1}=200$ years. Both terms have units of ppmv per year.

Thousands of years at constant $E_0$ erase the initial transient, givingThe reference $C_0$ is the zero-anthropogenic-emission equilibrium, not the equilibrium when the stated small persistent emissions are present.

Put $A=b_1{E}_0/b_0$. The proposed solution can be writtenIt gives $C(0)-C_0=A$. Differentiation and substitution give, respectively,so it satisfies both the differential equation and the initial condition. At $t=200$ years,with $E(2000)=0.02e^6\simeq 8.069\mathrm{GtC}\mathrm{y}{\mathrm{r}}^{-1}$. The earlier emissions grow on the $g^{-1}=33.3$ year timescale, much faster than the model's 200-year removal time.

Let $t=0$ now denote the year 2000 and $C_{00}=336.95\mathrm{ppmv}$. When emissions stop, the model givesThe concentration peaks at the switch, then returns exponentially to $279\mathrm{ppmv}$; in 2500 it is about $283.76\mathrm{ppmv}$. Before 2000 both scenarios share the exponential-emissions trajectory.

Using the specified logarithmic equilibrium warming relative to $C_0$,gives the fast-adjustment peak estimateAn unspecified finite thermal response time below twenty years delays and reduces that peak relative to the instantaneous-equilibrium value; its exact amount cannot be fixed without a response law and a chosen thermal time. The carbon-removal time is much longer, so the warming subsequently declines gradually in this model. Taking the 1800 concentration $280$ as the temperature reference instead changes the quoted estimate by only $0.016\mathrm{K}$.

Holding emissions at $E_{00}=8.069\mathrm{GtC}\mathrm{y}{\mathrm{r}}^{-1}$ gives equilibrium concentrationFor time $t$ since 2000,It increases monotonically on the same 200-year timescale and reaches about $654.07\mathrm{ppmv}$ in 2500. The limiting, and hence peak asymptotic, warming isIn 2500 the equilibrium warming is about $3.69\mathrm{K}$, still below that asymptote. A stable thermal adjustment much faster than the carbon timescale follows this slow approach with a short lag and has the same limiting warming. These are predictions of the specified linear reservoir model, not a claim that constant real emissions produce a finite universal atmospheric equilibrium.

The real carbon cycle has multiple reservoirs and adjustment times, rather than one universally exponential removal process. Air-sea exchange, ocean mixing and carbonate chemistry give a long-lived atmospheric tail; uptake also changes with temperature, circulation and carbon loading. For example, a warming ocean can absorb less additional carbon dioxide, so a fixed $b_0$ omits a potentially positive climate-carbon feedback. Terrestrial vegetation and soil responses provide further examples. One missing feature is the dependence of carbon uptake on the changing state of the ocean and climate.