A body cools from a temperature 3theta to 2th | vedclass

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Question

(C) According to Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\f

Vedclass · Accepted Answer

$\frac{3}{2}	heta$

Vedclass · Answer

(C) According to Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = -k(T - T_s)$.For the first interval of $10 	ext{ minutes}$:$\frac{3	heta - 2	heta}{10} = k \left( \frac{3	heta + 2	heta}{2} - 	heta ight)$$\frac{	heta}{10} = k(1.5	heta) = 1.5k	heta$$k = \frac{1}{15}$.For the next $10 	ext{ minutes}$, the body cools from $2	heta$ to $x$:$\frac{2	heta - x}{10} = k \left( \frac{2	heta + x}{2} - 	heta ight)$$\frac{2	heta - x}{10} = \frac{1}{15} \left( \frac{2	heta + x - 2	heta}{2} ight)$$\frac{2	heta - x}{10} = \frac{x}{30}$$3(2	heta - x) = x$$6	heta - 3x = x$$4x = 6	heta$$x = \frac{3}{2}	heta$.

$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be

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