If a liquid takes 30 ; s in cooling from 80^{ | vedclass

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(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} -

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$40$

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(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight)$,where $	heta_0$ is the room temperature.In the first case:$\frac{80 - 70}{30} = K \left( \frac{80 + 70}{2} - 	heta_0 ight) \Rightarrow \frac{10}{30} = K(75 - 	heta_0) \Rightarrow \frac{1}{3} = K(75 - 	heta_0)$  --- $(1)$In the second case:$\frac{60 - 50}{70} = K \left( \frac{60 + 50}{2} - 	heta_0 ight) \Rightarrow \frac{10}{70} = K(55 - 	heta_0) \Rightarrow \frac{1}{7} = K(55 - 	heta_0)$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{1/3}{1/7} = \frac{K(75 - 	heta_0)}{K(55 - 	heta_0)} \Rightarrow \frac{7}{3} = \frac{75 - 	heta_0}{55 - 	heta_0}$Cross-multiplying:$7(55 - 	heta_0) = 3(75 - 	heta_0)$$385 - 7	heta_0 = 225 - 3	heta_0$$385 - 225 = 7	heta_0 - 3	heta_0$$160 = 4	heta_0$$	heta_0 = 40^{\circ} C$.

If a liquid takes $30 \; s$ in cooling from $80^{\circ} C$ to $70^{\circ} C$ and $70 \; s$ in cooling from $60^{\circ} C$ to $50^{\circ} C$,then find the room temperature. (in $^{\circ} C$)

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