In Newton's law of cooling , dθdt = -kθ-θ0 , the constant k is proportional to

A , surface area of the body , 1 / m , m being mass of the body , e , emmisivity of the body

mc ( - d T d t ) = e A . 4 T 3 0 ( T - T 0 ) - d T d t = 4 T 3 0 A e m c T - T 0 , we find that K = A e m c K ∞ A , K ∞ e and K ∞ 1 m

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In Newton's law of cooling , $\frac{d θ}{d t} = - k (θ - θ_{0})$ , the constant k is proportional to

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$A$ , surface area of the body

$S$ , specific heat of the body

$1 / m, m$ being mass of the body

$e$ , emmisivity of the body

answer is A, C, D.

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mc $(- \frac{d T}{d t}) = e A . 4 {T^{3}}_{0} (T - T_{0})$

$(\begin{matrix} - d T \\ d t \end{matrix}) = \frac{4 {T^{3}}_{0} A e}{m c} (T - T_{0})$ , we find that

$K = \frac{A e}{m c}$

$K \infty A$ , K $\infty e$ and $K \infty \frac{1}{m}$

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