= Solution
The <heat capacity> per unit horizontal <area> is $C=\rho c_pd$: <mass density> times <specific heat capacity> times atmospheric depth. Thus $C\dot T$ is the rate of stored <thermal energy> per unit <area>. The <solar constant> $Q$ is the incident <radiative flux> on a surface normal to sunlight; the ratio of the Earth's intercepting disk <area> to its spherical <area> is $1/4$. The absorbed fraction is $1-a$, where <albedo> $a$ is the reflected fraction. The outgoing <radiative flux> is $\sigma\gamma T^4$, a <Stefan–Boltzmann law> with effective radiation factor $\gamma$ incorporating the simplified <greenhouse effect>. Every term therefore has units of <power> per <area>.
The <albedo> curve decreases smoothly, passes through $(T_0,1/2)$, and approaches $1/4$ at high <temperature>. At $T=0$ it is $(2+\tanh\mu)/4$, close to $3/4$ when $\mu$ is large. Its <derivative> is
$$
a_0'(T)=-\frac{\mu T^3}{T_0^4}\operatorname{sech}^2\!\left[\mu(T^4/T_0^4-1)\right]\leq0.
$$
An ice-covered planet reflects more sunlight than exposed land or ocean. Warming removes reflective <ice>, lowers <albedo>, and increases absorbed <radiative flux>: this is <ice-albedo feedback>. The parameter $\mu$ controls the sharpness of this transition; the formula is an idealized global response, rather than a local melting law.
\Image[/past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2010-climate-feedback.png]
{title=Temperature-dependent albedo and the separated fast cooling and slow recovery}
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