A body cools in 7 minutes from 60^circ C to 4 | vedclass

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(D) According to Newton's law of cooling, the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$, where $T$ is the temperature of the body, $T_

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$7$

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(D) According to Newton's law of cooling, the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$, where $T$ is the temperature of the body, $T_s$ is the surrounding temperature, and $k$ is a constant.For the first interval: $\frac{60 - 40}{7} = k \left( \frac{60 + 40}{2} - 10 \right) \implies \frac{20}{7} = k(50 - 10) \implies \frac{20}{7} = 40k \implies k = \frac{1}{14}$.For the second interval, let the time be $t$: $\frac{40 - 28}{t} = k \left( \frac{40 + 28}{2} - 10 \right) \implies \frac{12}{t} = \frac{1}{14} (34 - 10) \implies \frac{12}{t} = \frac{24}{14} \implies \frac{12}{t} = \frac{12}{7}$.Thus, $t = 7$ minutes.

$A$ body cools in $7$ minutes from $60^\circ C$ to $40^\circ C$. What time (in minutes) does it take to cool from $40^\circ C$ to $28^\circ C$, if its surrounding temperature is $10^\circ C$? (Newton's law of cooling holds good)

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