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

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

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$28$

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the temperature of the surroundings.For the first interval: $\frac{60 - 40}{6} = k \left( \frac{60 + 40}{2} - 10 \right) \implies \frac{20}{6} = k(50 - 10) \implies \frac{10}{3} = 40k \implies k = \frac{1}{12}$.For the second interval,let the final temperature be $T_f$: $\frac{40 - T_f}{6} = k \left( \frac{40 + T_f}{2} - 10 \right)$.Substituting $k = \frac{1}{12}$: $\frac{40 - T_f}{6} = \frac{1}{12} \left( 20 + \frac{T_f}{2} - 10 \right) \implies \frac{40 - T_f}{6} = \frac{1}{12} \left( 10 + \frac{T_f}{2} \right)$.Multiplying by $12$: $2(40 - T_f) = 10 + 0.5T_f \implies 80 - 2T_f = 10 + 0.5T_f \implies 70 = 2.5T_f \implies T_f = \frac{70}{2.5} = 28^{\circ} C$.

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

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