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

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(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 interval: $\f

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 interval: $\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 25 ight)$.$1 = K(55 - 25) \Rightarrow 1 = 30K \Rightarrow K = \frac{1}{30}$.For the next interval,let the final temperature be $	heta_f$. Then $\frac{50 - 	heta_f}{10} = K \left( \frac{50 + 	heta_f}{2} - 25 ight)$.Substituting $K = \frac{1}{30}$: $\frac{50 - 	heta_f}{10} = \frac{1}{30} \left( \frac{50 + 	heta_f - 50}{2} ight)$.$\frac{50 - 	heta_f}{10} = \frac{	heta_f}{60}$.$6(50 - 	heta_f) = 	heta_f$.$300 - 6	heta_f = 	heta_f$.$7	heta_f = 300 \Rightarrow 	heta_f = \frac{300}{7} \approx 42.85\,^oC$.

$A$ 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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