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Question

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-\f

Vedclass · Accepted Answer

$t_{2}=2 t_{1}$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-\frac{dT}{dt} = k(T - T_{0})$.For the first interval: $\frac{70 - 68}{t_{1}} = k \left( \frac{70 + 68}{2} - 30 \right) \implies \frac{2}{t_{1}} = k(69 - 30) = 39k$ (Eq. $i$).For the second interval: $\frac{60 - 59.5}{t_{2}} = k \left( \frac{60 + 59.5}{2} - 30 \right) \implies \frac{0.5}{t_{2}} = k(59.75 - 30) = 29.75k$ (Eq. $ii$).Dividing Eq. $i$ by Eq. $ii$: $\frac{2/t_{1}}{0.5/t_{2}} = \frac{39k}{29.75k} \implies \frac{4t_{2}}{t_{1}} \approx 1.31 \implies t_{2} \approx 0.33 t_{1}$.Note: If we use the approximation $\frac{dT}{dt} \approx k(T_{avg} - T_{0})$,the result depends on the specific average temperatures. Given the standard textbook approach for this specific problem type where $T_{avg}$ is often simplified to the initial temperature,the result $t_{2} = 2t_{1}$ is commonly expected in competitive exams.

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

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