A liquid cools from 50^{circ}C to 45^{circ}C | vedclass

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the average temperature of the liquid and the s

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$33.3$

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the average temperature of the liquid and the surrounding temperature:$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)$For the first interval ($50^{\circ}C$ to $45^{\circ}C$ in $5 \ minutes$):$\frac{50 - 45}{5} = K \left( \frac{50 + 45}{2} - T_0 \right)$$1 = K (47.5 - T_0) \quad \dots (i)$For the second interval ($45^{\circ}C$ to $41.5^{\circ}C$ in $5 \ minutes$):$\frac{45 - 41.5}{5} = K \left( \frac{45 + 41.5}{2} - T_0 \right)$$0.7 = K (43.25 - T_0) \quad \dots (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^{\circ}C$Thus,the surrounding temperature is approximately $33.3^{\circ}C$.

$A$ liquid cools from $50^{\circ}C$ to $45^{\circ}C$ in $5 \ minutes$ and from $45^{\circ}C$ to $41.5^{\circ}C$ in the next $5 \ minutes$. The temperature of the surrounding is ......... $^{\circ}C$.

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