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,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.For the first $10$ minutes: $\frac{60

Vedclass · Accepted Answer

$42.85$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.For the first $10$ minutes: $\frac{60 - 50}{10} = K\left( \frac{60 + 50}{2} - 25 ight) \Rightarrow 1 = K(55 - 25) \Rightarrow 1 = 30K$  --- $(i)$For the next $10$ minutes,let the final temperature be $	heta$: $\frac{50 - 	heta}{10} = K\left( \frac{50 + 	heta}{2} - 25 ight) \Rightarrow \frac{50 - 	heta}{10} = K\left( \frac{50 + 	heta - 50}{2} ight) \Rightarrow \frac{50 - 	heta}{10} = K\left( \frac{	heta}{2} ight)$ --- $(ii)$Dividing $(i)$ by $(ii)$: $\frac{1}{(50 - 	heta)/10} = \frac{30K}{K(	heta/2)} \Rightarrow \frac{10}{50 - 	heta} = \frac{60}{	heta}$.Cross-multiplying: $10	heta = 60(50 - 	heta) \Rightarrow 10	heta = 3000 - 60	heta \Rightarrow 70	heta = 3000$.$	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 assuming Newton's law of cooling to hold good,the temperature of the body at the end of the next $10$ minutes will be ......... $^{\circ}C$.

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$A$ 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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