A metal sphere cools at the rate of 1.5^{circ | vedclass

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Question

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\fr

Vedclass · Accepted Answer

$0.6^{\circ} C / min$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = k(T - T_s)$.For the first case,$T_1 = 80^{\circ} C$ and $T_s = 30^{\circ} C$. The rate is $1.5^{\circ} C / min$.$1.5 = k(80 - 30) = k(50) \implies k = \frac{1.5}{50} = 0.03 \ min^{-1}$.For the second case,$T_2 = 50^{\circ} C$ and $T_s = 30^{\circ} C$. Let the rate be $r$.$r = k(50 - 30) = k(20)$.Substituting the value of $k$:$r = 0.03 	imes 20 = 0.6^{\circ} C / min$.

$A$ metal sphere cools at the rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. At what rate will it cool when its temperature falls to $50^{\circ} C$? [Temperature of surrounding is $30^{\circ} C$]

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