A body cools from 60^{circ} C to 40^{circ} C | vedclass

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

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$28$

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(A) According to Newton's law of cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $\frac{dT}{dt} = -k(T - T_s)$.For the first interval of $6$ minutes:$\frac{60 - 40}{6} = k \left( \frac{60 + 40}{2} - 10 \right)$$\frac{20}{6} = k(50 - 10) = 40k$  --- $(i)$For the next $6$ minutes,let the final temperature be $T$:$\frac{40 - T}{6} = k \left( \frac{40 + T}{2} - 10 \right)$$\frac{40 - T}{6} = k \left( \frac{40 + T - 20}{2} \right) = k \left( \frac{20 + T}{2} \right)$ --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{20 / 6}{(40 - T) / 6} = \frac{40k}{k(20 + T) / 2}$$\frac{20}{40 - T} = \frac{80}{20 + T}$$20(20 + T) = 80(40 - T)$$400 + 20T = 3200 - 80T$$100T = 2800$$T = 28^{\circ} C$.

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. If the temperature of the surroundings is $10^{\circ} C$,then after the next $6$ minutes,its temperature will be $.........{ }^{\circ} C$.

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