A metallic sphere cools from 50^{circ}C to 40 | vedclass

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the surrounding temperature.For

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$33$

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the surrounding temperature.For the first interval: $\frac{50 - 40}{300} = k \left( \frac{50 + 40}{2} - 20 \right)$.$\frac{10}{300} = k(45 - 20) \implies \frac{1}{30} = 25k \implies k = \frac{1}{750}$.For the next $5$ minutes $(300 \, s)$,let the final temperature be $T$. Then: $\frac{40 - T}{300} = k \left( \frac{40 + T}{2} - 20 \right)$.Substituting $k = \frac{1}{750}$: $\frac{40 - T}{300} = \frac{1}{750} \left( \frac{40 + T - 40}{2} \right)$.$\frac{40 - T}{300} = \frac{T}{1500}$.$5(40 - T) = T \implies 200 - 5T = T \implies 6T = 200$.$T = \frac{200}{6} \approx 33.33^{\circ}C$. The closest integer value is $33^{\circ}C$.

$A$ metallic sphere cools from $50^{\circ}C$ to $40^{\circ}C$ in $300 \, s$. If the atmospheric temperature is $20^{\circ}C$,then the sphere's temperature after the next $5$ minutes will be close to $.....^{\circ}C$.

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