A body takes T minutes to cool from 62^{circ} | vedclass

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_s)$,where $	heta_{avg}$ is the

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Greater than $T$ minutes

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(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_s)$,where $	heta_{avg}$ is the average temperature of the body and $	heta_s$ is the surrounding temperature.For the first case: $\frac{62 - 61}{T} = K \left( \frac{62 + 61}{2} - 30 ight) \implies \frac{1}{T} = K(61.5 - 30) = K(31.5)$ ...$(i)$For the second case,cooling from $46^{\circ}C$ to $45^{\circ}C$ in time $T'$: $\frac{46 - 45}{T'} = K \left( \frac{46 + 45}{2} - 30 ight) \implies \frac{1}{T'} = K(45.5 - 30) = K(15.5)$ ...$(ii)$Comparing $(i)$ and $(ii)$,since $31.5 > 15.5$,the rate of cooling is faster in the first case. Therefore,the time taken $T'$ for the second case must be greater than $T$.

$A$ body takes $T$ minutes to cool from $62^{\circ}C$ to $61^{\circ}C$ when the surrounding temperature is $30^{\circ}C$. The time taken by the body to cool from $46^{\circ}C$ to $45^{\circ}C$ is:

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