A body cools from 60,^oC to 50,^oC in 10,minu | vedclass

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Question

(C) According to Newton's law of cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$.For the first $10\,minutes

Vedclass · Accepted Answer

$42.85$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$.For the first $10\,minutes$:$\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 25 ight)$$1 = K(55 - 25) \Rightarrow 1 = 30K \Rightarrow K = \frac{1}{30}$.For the next $10\,minutes$,let the final temperature be $	heta$:$\frac{50 - 	heta}{10} = K \left( \frac{50 + 	heta}{2} - 25 ight)$$\frac{50 - 	heta}{10} = \frac{1}{30} \left( \frac{50 + 	heta - 50}{2} ight)$$\frac{50 - 	heta}{10} = \frac{1}{30} \left( \frac{	heta}{2} ight)$$3(50 - 	heta) = \frac{	heta}{2}$$150 - 3	heta = 0.5	heta$$3.5	heta = 150$$	heta = \frac{150}{3.5} = \frac{300}{7} \approx 42.85\,^oC$.

$A$ body cools from $60\,^oC$ to $50\,^oC$ in $10\,minutes$. If the room temperature is $25\,^oC$ 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 ......... $^oC$.

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