A body takes 10 minutes to cool from 60^{cir | vedclass

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(C) 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 and $

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(C) 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 and $T_s$ is the temperature of the surroundings.For the first interval: $\frac{60 - 50}{10} = k \left( \frac{60 + 50}{2} - 25 \right) \implies 1 = k(55 - 25) \implies 1 = 30k \implies k = \frac{1}{30}$.For the second interval,let the final temperature be $T_f$: $\frac{50 - T_f}{10} = k \left( \frac{50 + T_f}{2} - 25 \right)$.Substituting $k = \frac{1}{30}$: $\frac{50 - T_f}{10} = \frac{1}{30} \left( \frac{50 + T_f - 50}{2} \right) \implies \frac{50 - T_f}{10} = \frac{T_f}{60}$.Multiplying by $60$: $6(50 - T_f) = T_f \implies 300 - 6T_f = T_f \implies 7T_f = 300 \implies T_f = \frac{300}{7} \approx 42.86^{\circ}C$.Rounding to the nearest integer provided in the options,the value is $43^{\circ}C$.

$A$ body takes $10 \ minutes$ to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the same body after the next $10 \ minutes$ will be (if the temperature of the surroundings is $25^{\circ}C$): (in $^{\circ}C$)

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