A body cools from 60^{circ}C to 50^{circ}C in | vedclass

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(C) According to Newton's law of cooling,$\frac{dT}{dt} = -K(T - T_{room})$.For the first interval: $\frac{60 - 50}{10} = K\left( \frac{60 + 50}{2} -

Vedclass · Accepted Answer

$42.85$

Vedclass · Answer

(C) According to Newton's law of cooling,$\frac{dT}{dt} = -K(T - T_{room})$.For the first interval: $\frac{60 - 50}{10} = K\left( \frac{60 + 50}{2} - 25 ight) \implies 1 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30}$.For the second interval,let the final temperature be $	heta$: $\frac{50 - 	heta}{10} = K\left( \frac{50 + 	heta}{2} - 25 ight)$.Substituting $K = \frac{1}{30}$: $\frac{50 - 	heta}{10} = \frac{1}{30} \left( \frac{50 + 	heta - 50}{2} ight) = \frac{1}{30} \left( \frac{	heta}{2} ight) = \frac{	heta}{60}$.$6(50 - 	heta) = 	heta \implies 300 - 6	heta = 	heta \implies 7	heta = 300 \implies 	heta = \frac{300}{7} \approx 42.85^{\circ}C$.

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and Newton's law of cooling is applicable,what will be the temperature of the body after another $10$ minutes in $^{\circ}C$?

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