The initial temperature of a body is 80^{circ | vedclass

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(B) According to Newton's law of cooling,the rate of change of temperature is given by $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	he

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of change of temperature 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 surrounding temperature.For the first interval ($t = 5$ minutes):$\frac{80 - 64}{5} = K \left[ \frac{80 + 64}{2} - 	heta_0 ight] \implies 3.2 = K(72 - 	heta_0)$ ... $(i)$For the second interval ($t = 10$ minutes,from $64^{\circ}C$ to $52^{\circ}C$):$\frac{64 - 52}{10} = K \left[ \frac{64 + 52}{2} - 	heta_0 ight] \implies 1.2 = K(58 - 	heta_0)$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{3.2}{1.2} = \frac{72 - 	heta_0}{58 - 	heta_0} \implies \frac{8}{3} = \frac{72 - 	heta_0}{58 - 	heta_0}$$8(58 - 	heta_0) = 3(72 - 	heta_0) \implies 464 - 8	heta_0 = 216 - 3	heta_0$$5	heta_0 = 248 \implies 	heta_0 = 49.6^{\circ}C$. Given the options,the closest value is $49^{\circ}C$.

The initial temperature of a body is $80^{\circ}C$. If its temperature falls to $64^{\circ}C$ in $5$ minutes and in $10$ minutes to $52^{\circ}C$,then the temperature of the surrounding will be ...... $^{\circ}C$.

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