 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s25/3/1/solution

Put

ϑ = T - T_{0}

and assume

C > 0

λ > 0

\dot{ϑ} + \frac{λ}{C} ϑ = \frac{F _{0}}{C}, ϑ (0) = 0.

e^{λ t / C}

and integrating from zero to

t

gives

T (t) = T_{0} + \frac{F _{0}}{λ} (1 - e^{- k_{T} t}), k_{T} = \frac{λ}{C} .

The response initially grows at rate

F_{0} / C

and tends to the equilibrium warming

F_{0} / λ

. Heat capacity per unit area and radiative-feedback units make

k_{T}

an inverse time.

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