Use a quasistatic surface energy balance with negligible heat storage in the surface or ice, a linear ice temperature profile, and a fixed freezing temperature at the ice-ocean interface. These are the assumptions behind the <single-category sea-ice thermodynamic model> used below. Write $\theta=t/\tau$, $F_O=\lambda_O(T_O-T_m)>0$ and
$$
r=\frac{F_O}{\lambda_A\Delta T},\qquad a=\frac{\lambda_A}{k},\qquad
H=\left(\frac{\kappa\tau}{\mathcal S}\right)^{1/2},\qquad
\mathcal S=\frac{L}{c_p\Delta T}.
$$
Here $\lambda_A$ and $\lambda_O$ are <heat transfer coefficients>, $k$ is <thermal conductivity>, $\kappa=k/(\rho c_p)$ is <thermal diffusivity>, and $\mathcal S$ is the <inverse Stefan number>. Here $L$ is <latent heat> per unit mass. The thermal $k$ denotes conductivity; the previous questions use $k$ for permeability.

While there is open water, the upward ocean heat supply balances the loss to air: $\lambda_O(T_O-T_s)=\lambda_A(T_s-T_A)$. Thus
$$
\boxed{T_s^{\rm water}=\frac{\lambda_AT_A+\lambda_OT_O}{\lambda_A+\lambda_O}}.
$$
Freezing begins when this surface reaches $T_m$ on the cooling part of the annual cycle, which requires $r<1$ and gives
$$
\boxed{t_1=\tau\arccos(-r),\qquad\frac\pi2<\frac{t_1}{\tau}<\pi}.
$$
For $r\ge1$ the surface never cools strictly below freezing and no positive ice thickness develops in this model. The limit $r\to0$ gives $t_1\to\pi\tau/2$.

With ice present and a cold surface, <thermal conduction> and <Newton's law of cooling> give $k(T_m-T_s)/h=\lambda_A(T_s-T_A)$. The surface temperature is
$$
\boxed{T_s^{\rm ice}=\frac{kT_m+\lambda_AhT_A}{k+\lambda_Ah}\quad(T_A<T_m)}.
$$
During surface melting the temperature is capped at $T_s=T_m$; the ice is not permitted to follow the above formula to a temperature higher than its melting point. The upward conductive <heat flux> during the cold phase is
$$
F_C=\frac{\lambda_A\Delta T(-\cos\theta)}{1+ah}.
$$
The <Stefan condition> at the ocean interface gives $\rho L\dot h=F_C-F_O$. Under the permitted approximation of negligible ocean heat during growth,
$$
\dot h=\frac{\kappa}{\mathcal S}\frac{-\cos(t/\tau)}{h+1/a}.
$$
Starting from $h(t_1)=0$, integration yields the implicit growth law
$$
\boxed{\frac{h(t)^2}{2}+\frac{k}{\lambda_A}h(t)
=\frac{\kappa\tau}{\mathcal S}\left[\sin(t_1/\tau)-\sin(t/\tau)\right]}.
$$
The leading maximum occurs when the air returns to freezing from below:
$$
\boxed{t_2\simeq\frac{3\pi}{2}\tau,\qquad
h_m=\left[\left(\frac{k}{\lambda_A}\right)^2+\frac{2\kappa\tau}{\mathcal S}(1+\sin(t_1/\tau))\right]^{1/2}-\frac{k}{\lambda_A}}.
$$
The equality for $h_m$ is within the ocean-heat-neglected growth model, with the freezing time set by the open-water balance. If $aH\gg1$, conductive resistance through the ice dominates the atmospheric boundary resistance at maximum thickness. Consequently
$$
\boxed{h_m\sim[1+\sin(t_1/\tau)]^{1/2}\left(\frac{2\kappa\tau}{\mathcal S}\right)^{1/2}}.
$$
This reproduces the displayed large-conductance limit. A sufficient condition for the quasistatic temperature assumption is $h_m^2/\kappa\ll\tau$; large $\mathcal S$ ensures this in the thick-ice scaling.

The approximation concerning ocean heat needs care. Its ratio to atmospheric heat removal during the cold phase is
$$
\frac{F_O}{F_C}=\frac{r(1+ah)}{-\cos\theta}.
$$
It is uniformly small away from the beginnings and ends of winter provided
$$
\boxed{r(1+ah_m)\ll1,\quad\text{that is,}\quad
\lambda_O(T_O-T_m)\ll\frac{\lambda_A\Delta T}{1+\lambda_Ah_m/k}}.
$$
For the thick-ice limit this requires both $r\ll1$ and $r\,aH\ll1$. This also makes the integrated ocean heat small relative to the latent heat of the maximum ice column. It is not possible to neglect $F_O$ pointwise throughout the closed growth interval: at the actual freezing onset it balances the atmospheric removal, and at the true maximum the growth rate is zero because the two fluxes again balance. The <ocean-heat correction to seasonal ice growth> therefore gives narrow endpoint corrections. In the full model the maximum is slightly earlier than $3\pi\tau/2$, with
$$
-\cos(t_2/\tau)=r[1+ah(t_2)].
$$
Thus $t_2=3\pi\tau/2-O(\tau r[1+ah_m])$ in the stated approximation, rather than an exact maximum time for finite ocean heat.

For $3\pi\tau/2\le t\le2\pi\tau$, while ice survives, both the warm air and ocean melt it. The surface is at $T_m$, and the total <Stefan condition> is
$$
\rho L\dot h=-\lambda_A\Delta T\cos(t/\tau)-F_O.
$$
Integrating from the leading maximum gives
$$
h(t)=h_m-\frac{\lambda_A\Delta T\tau}{\rho L}\left[1+\sin(t/\tau)+r\left(\frac t\tau-\frac{3\pi}{2}\right)\right]
$$
until its first zero. The physical thickness is then zero for the rest of this warm interval and the open-water temperature applies. Therefore at the end of the first year,
$$
\boxed{h(2\pi\tau)\simeq\max\left\{0,h_m-\frac{\lambda_A\kappa\tau}{k\mathcal S}\left[1+\frac\pi2\frac{\lambda_O}{\lambda_A}\frac{T_O-T_m}{\Delta T}\right]\right\}}.
$$
Positive ice at that time is equivalent, in the specified approximation, to
$$
\boxed{\frac{\lambda_A\kappa\tau}{k\mathcal S}\left[1+\frac\pi2\frac{\lambda_O}{\lambda_A}\frac{T_O-T_m}{\Delta T}\right]<h_m}.
$$
This establishes the algebraic criterion printed in the paper.

There is, however, an actual qualification to the paper's word “perennial.” This inequality tests <peak-temperature sea-ice survival>, since $t=2\pi\tau$ is the next atmospheric temperature maximum. Surface melting continues until at least $t=5\pi\tau/2$, so positive thickness at the maximum does not guarantee survival throughout the summer. In the same approximation, survival through that full warm half-cycle would require the stronger inequality
$$
h_m>\frac{\lambda_A\Delta T\tau}{\rho L}(2+\pi r).
$$
In fact, starting ice-free under the specified zero-mean sinusoidal forcing, even the zero-ocean-heat upper bound on winter growth cannot meet it. In that limit $t_1=\pi\tau/2$ and
$$
\frac{h_m^2}{2}+\frac{h_m}{a}=2H^2,
\qquad
h_m<2aH^2=\frac{2\lambda_A\Delta T\tau}{\rho L},
$$
since $h_m^2/2>0$. The expression on the right is already the atmospheric melt over the entire subsequent warm half-cycle. Positive ocean heat can only reduce the winter ice and increase summer melt. Thus the printed condition is a first-year peak-temperature survival test, not a sufficient criterion for literal year-round ice in this model. For example, with $aH=1/2$ and very small positive $r$, it gives $h_m/H\simeq2(\sqrt2-1)\simeq0.828$, while first-year melt is approximately $0.5H$ but full warm-season melt is approximately $H$. Ice is present at the year's endpoint and disappears later that same summer. No reinterpretation of the printed inequality as guaranteed perennial cover is justified without an additional seasonal assumption.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 332 / 4 / Solution

Put $\Delta=T_0-T_\infty$. Before either forcing term acts, <Newton's law of cooling> gives
$$
\boxed{T_1(t)=T_\infty+\Delta e^{-at}\quad(0<t\leq1)},
\qquad
\boxed{T_2(t)=T_\infty+\Delta e^{-at}\quad(t>0)}.
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 7C / a / Solution

{c}
{wiki}

= Newton law of cooling
{c}
{synonym}

Newton's law of cooling models the temperature $T$ of an object in an environment at constant temperature $T_\infty$ by $\dot T=-a(T-T_\infty)$ for a positive cooling rate $a$.


Codex @codex  0

 Incoming links: Newton's law of cooling