= Solution
Solar geometry must be included before multiplying by the summer duration. Let $\phi=71^\circ$, solar declination $\delta$, and hour angle $u$, measured from local noon. The cosine of solar zenith angle is
$$
\cos\zeta=\sin\phi\sin\delta+\cos\phi\cos\delta\cos u.
$$
Only positive values receive sunlight. Integrating through a day gives the <daily mean solar irradiance> at the top of the atmosphere:
$$
\overline S=\frac{S_0}{\pi}
[H_0\sin\phi\sin\delta+\cos\phi\cos\delta\sin H_0],
\qquad H_0=\arccos(-\tan\phi\tan\delta),
$$
with $H_0$ clipped to $\pi$ for polar day and to zero for polar night. At the solstice, for example, polar-day averaging gives $\overline S=S_0\sin71^\circ\sin23.44^\circ\simeq514\,\mathrm{W\,m^{-2}}$. 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 $427\,\mathrm{W\,m^{-2}}$. There are 92 days. With <surface albedo> $\alpha=0.1$, the no-atmosphere absorbed-solar ceiling is
$$
\boxed{Q_{\rm solar,TOA}
=(1-\alpha)\sum_{n=152}^{243}\overline S(n)(86400)
\simeq3.06\times10^9\,\mathrm{J\,m^{-2}}.}
$$
This calculation neglects the small seasonal change in Earth-Sun distance. The ceiling is larger than the $1.80\,\mathrm{GJ\,m^{-2}}$ 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 $\overline L$:
$$
Q_{\rm stored}\simeq\tau Q_{\rm solar,TOA}
-\overline L(92)(86400),
$$
before adding advection or subtracting ice melting. The parameters are scenario assumptions, not measurements supplied by the question. For example, $\tau=0.6$ gives $1.83\,\mathrm{GJ\,m^{-2}}$ before other losses, barely enough; with a modest mean loss of $50\,\mathrm{W\,m^{-2}}$, the retained amount is only $1.44\,\mathrm{GJ\,m^{-2}}$. 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 $1.804/3.058\simeq0.590$; 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.
\b[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 $0.361\,\mathrm{GJ\,m^{-2}}$, 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.
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