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

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(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K(	heta - 	heta_0)$,where $	heta$ is the temperatur

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$30$

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(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K(	heta - 	heta_0)$,where $	heta$ is the temperature of the body and $	heta_0$ is the temperature of the surroundings.For the first case:$1.5 = K(80 - 	heta_0)$  --- (Equation $1$)For the second case:$0.3 = K(40 - 	heta_0)$  --- (Equation $2$)Dividing Equation $1$ by Equation $2$:$\frac{1.5}{0.3} = \frac{K(80 - 	heta_0)}{K(40 - 	heta_0)}$$5 = \frac{80 - 	heta_0}{40 - 	heta_0}$$5(40 - 	heta_0) = 80 - 	heta_0$$200 - 5	heta_0 = 80 - 	heta_0$$200 - 80 = 5	heta_0 - 	heta_0$$120 = 4	heta_0$$	heta_0 = 30^{\circ} C$Thus,the temperature of the surrounding is $30^{\circ} C$.

$A$ metal sphere cools at a rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. When the temperature of the sphere is $40^{\circ} C$,its rate of cooling is $0.3^{\circ} C / min$. The temperature of the surrounding $\left(\theta_0\right)$ is (in $^{\circ} C$)

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