An ordinary body cools from 4 theta to 3 thet | vedclass

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(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$,where $	heta_0$ is the sur

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$\frac{7 	heta}{3}$

Vedclass · Answer

(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$,where $	heta_0$ is the surrounding temperature.For the first interval:$\frac{4	heta - 3	heta}{t} = K \left( \frac{4	heta + 3	heta}{2} - 	heta ight)$$\frac{	heta}{t} = K \left( \frac{7	heta}{2} - 	heta ight) = K \left( \frac{5	heta}{2} ight)$$K = \frac{2}{5t} \dots (i)$For the next interval of $t$ minutes,let the final temperature be $x$:$\frac{3	heta - x}{t} = K \left( \frac{3	heta + x}{2} - 	heta ight)$Substituting $K$ from $(i)$:$\frac{3	heta - x}{t} = \frac{2}{5t} \left( \frac{3	heta + x - 2	heta}{2} ight)$$3	heta - x = \frac{1}{5} (	heta + x)$$15	heta - 5x = 	heta + x$$6x = 14	heta$$x = \frac{14	heta}{6} = \frac{7	heta}{3}$

An ordinary body cools from $4 \theta$ to $3 \theta$ in $t$ minutes. The temperature of that body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature is $\theta$)

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