In 5, text{minutes},a body cools from 75^{cir | vedclass

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(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T_{avg} - T_s)$.For the first $5\, 	ext{minutes}$:$\frac

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$57$

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(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T_{avg} - T_s)$.For the first $5\, 	ext{minutes}$:$\frac{75 - 65}{5} = k \left( \frac{75 + 65}{2} - 25 ight)$$\frac{10}{5} = k (70 - 25)$$2 = k(45) \implies k = \frac{2}{45} \, 	ext{min}^{-1}$.For the next $5\, 	ext{minutes}$,let the final temperature be $T$:$\frac{65 - T}{5} = k \left( \frac{65 + T}{2} - 25 ight)$$\frac{65 - T}{5} = \frac{2}{45} \left( \frac{65 + T - 50}{2} ight)$$\frac{65 - T}{5} = \frac{1}{45} (T + 15)$$9(65 - T) = T + 15$$585 - 9T = T + 15$$10T = 570$$T = 57^{\circ} 	ext{C}$.

In $5\, \text{minutes}$,a body cools from $75^{\circ} \text{C}$ to $65^{\circ} \text{C}$ at a room temperature of $25^{\circ} \text{C}$. The temperature of the body at the end of the next $5\, \text{minutes}$ is $......\,^{\circ} \text{C}$.

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