Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 72 / 4 / b / ii / Solution

Solar geometry must be included before multiplying by the summer duration. Let $\varphi =71^{\circ }$, solar declination $\delta$, and hour angle $u$, measured from local noon. The cosine of solar zenith angle isOnly positive values receive sunlight. Integrating through a day gives the daily mean solar irradiance at the top of the atmosphere:with $H_0$ clipped to $\pi$ for polar day and to zero for polar night. At the solstice, for example, polar-day averaging gives $\overline{S}=S_0\sin 71^{\circ }\sin 23.44^{\circ }\simeq 514\mathrm{W}{\mathrm{m}}^{-2}$. It would be wrong to apply $S_0$ continuously to a horizontal surface.

A simple seasonal approximation $\delta (n)=23.44^{\circ }\sin [2\pi (n-80)/365]$, with calendar day $n$, gives a June–August daily-mean average of about $427\mathrm{W}{\mathrm{m}}^{-2}$. There are 92 days. With surface albedo $\alpha =0.1$, the no-atmosphere absorbed-solar ceiling isThis calculation neglects the small seasonal change in Earth-Sun distance. The ceiling is larger than the $1.80\mathrm{GJ}{\mathrm{m}}^{-2}$ required in part (i), so geometry alone does not make a uniform $7^{\circ }\mathrm{C}$ column impossible.

A reasonable conditional estimate includes an effective atmospheric short-wave transmission $\tau$ and net non-solar loss $\overline{L}$:before adding advection or subtracting ice melting. The parameters are scenario assumptions, not measurements supplied by the question. For example, $\tau =0.6$ gives $1.83\mathrm{GJ}{\mathrm{m}}^{-2}$ before other losses, barely enough; with a modest mean loss of $50\mathrm{W}{\mathrm{m}}^{-2}$, the retained amount is only $1.44\mathrm{GJ}{\mathrm{m}}^{-2}$. That would raise a uniform 50 m column from freezing by about $7.0\mathrm{K}$, reaching roughly $5.2^{\circ }\mathrm{C}$. With no other losses the transmission required for $7^{\circ }\mathrm{C}$ is $1.804/3.058\simeq 0.590$; with that illustrative loss it rises to about $0.720$. Clouds, emitted thermal radiation, evaporation, transfer to colder water and melting all affect the balance.

The satellite surface temperature is insufficient evidence for a $7^{\circ }\mathrm{C}$ seabed. A warm, shallow ocean mixed layer can overlie colder water because meltwater and salinity maintain stable density stratification. Heating only the upper 10 m through $8.8\mathrm{K}$ costs $0.361\mathrm{GJ}{\mathrm{m}}^{-2}$, much less than heating all 50 m. Warm Pacific-water advection can also raise surface temperature or supply additional heat. If measured net solar input were too small for full-depth warming, shallow surface heating would be the natural alternative; if mixing and additional heat supply were strong enough, full-depth warming remains possible. The missing transmission, loss, mixing and inflow information prevents a unique yes-or-no conclusion from the supplied surface observation.