A body cools from a temperature 60,^{circ}tex | 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 $T$ is the temperature of the body,$T_s$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) 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: $\frac{60 - 40}{10} = k \left( \frac{60 + 40}{2} - 20 ight)$.$2 = k(50 - 20) = 30k$,so $k = \frac{2}{30} = \frac{1}{15}$.For the next interval of $10$ minutes,let the final temperature be $T_f$.$\frac{40 - T_f}{10} = k \left( \frac{40 + T_f}{2} - 20 ight)$.Substituting $k = \frac{1}{15}$: $\frac{40 - T_f}{10} = \frac{1}{15} \left( \frac{40 + T_f - 40}{2} ight) = \frac{1}{15} \left( \frac{T_f}{2} ight) = \frac{T_f}{30}$.Multiplying by $30$: $3(40 - T_f) = T_f$.$120 - 3T_f = T_f \implies 4T_f = 120 \implies T_f = 30\,^{\circ}	ext{C}$.

$A$ body cools from a temperature $60\,^{\circ}\text{C}$ to $40\,^{\circ}\text{C}$ in $10$ minutes. The room temperature is $20\,^{\circ}\text{C}$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be......... $^{\circ}\text{C}$.

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