A liquid cools from 50^oC to 45^oC in 5 minut | vedclass

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Question

(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding

Vedclass · Accepted Answer

$33.3$

Vedclass · Answer

(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T_{avg} - T_0)$.For the first interval: $\frac{50 - 45}{5} = K(\frac{50 + 45}{2} - T_0) \implies 1 = K(47.5 - T_0)$ ... $(i)$For the second interval: $\frac{45 - 41.5}{5} = K(\frac{45 + 41.5}{2} - T_0) \implies 0.7 = K(43.25 - T_0)$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{1}{0.7} = \frac{47.5 - T_0}{43.25 - T_0}$$43.25 - T_0 = 0.7(47.5 - T_0)$$43.25 - T_0 = 33.25 - 0.7T_0$$0.3T_0 = 10$$T_0 = \frac{10}{0.3} = 33.33\;^oC$.Thus,the surrounding temperature is $33.3\;^oC$.

$A$ liquid cools from $50^oC$ to $45^oC$ in $5$ minutes and from $45^oC$ to $41.5^oC$ in the next $5$ minutes. The temperature of the surrounding is ...... $^oC$.

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