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(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and the surroundings: $

Vedclass · Accepted Answer

$33.3$

Vedclass · Answer

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and the surroundings: $\frac{dT}{dt} = K(T_{avg} - T_s)$.In the first $10$ minutes,the temperature falls from $50^oC$ to $40^oC$ with surroundings at $20^oC$:$\frac{50 - 40}{10} = K \left( \frac{50 + 40}{2} - 20 ight) \implies 1 = K(45 - 20) \implies 1 = 25K \implies K = \frac{1}{25} \dots (i)$In the next $10$ minutes,let the final temperature be $	heta_2$:$\frac{40 - 	heta_2}{10} = K \left( \frac{40 + 	heta_2}{2} - 20 ight) \dots (ii)$Substituting $K = \frac{1}{25}$ into equation $(ii)$:$\frac{40 - 	heta_2}{10} = \frac{1}{25} \left( 20 + \frac{	heta_2}{2} - 20 ight) \implies \frac{40 - 	heta_2}{10} = \frac{	heta_2}{50}$$5(40 - 	heta_2) = 	heta_2 \implies 200 - 5	heta_2 = 	heta_2 \implies 6	heta_2 = 200$$	heta_2 = \frac{200}{6} = 33.33^oC$.

The temperature of a body falls from $50^oC$ to $40^oC$ in $10$ minutes. If the temperature of the surroundings is $20^oC$,then the temperature of the body after another $10$ minutes will be ........ $^oC$.

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