A metal rod cools at the rate of 4^{circ}C/mi | vedclass

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(C) According to Newton's law of cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surro

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$10$

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(C) According to Newton's law of cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surroundings: $R = k(	heta - 	heta_0)$.Given:Case $1$: $R_1 = 4^{\circ}C/min$,$	heta_1 = 90^{\circ}C$Case $2$: $R_2 = 1^{\circ}C/min$,$	heta_2 = 30^{\circ}C$Taking the ratio of the two rates:$\frac{R_1}{R_2} = \frac{	heta_1 - 	heta_0}{	heta_2 - 	heta_0}$$\frac{4}{1} = \frac{90 - 	heta_0}{30 - 	heta_0}$$4(30 - 	heta_0) = 90 - 	heta_0$$120 - 4	heta_0 = 90 - 	heta_0$$120 - 90 = 4	heta_0 - 	heta_0$$30 = 3	heta_0$$	heta_0 = 10^{\circ}C$Thus,the temperature of the surroundings is $10^{\circ}C$.

$A$ metal rod cools at the rate of $4^{\circ}C/min$ when its temperature is $90^{\circ}C$ and at the rate of $1^{\circ}C/min$ when its temperature is $30^{\circ}C$. The temperature of the surrounding is: (in $^{\circ}C$)

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