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(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $(T - T_s) = \Delta T$ is the temperature

Vedclass · Accepted Answer

$\frac{\ln(2) \cdot \Delta T_0}{R}$

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(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $(T - T_s) = \Delta T$ is the temperature difference.Given the initial rate of cooling is $R = k \Delta T_0$,we can find the rate constant $k = \frac{R}{\Delta T_0}$.The temperature difference at any time $t$ is given by $\Delta T(t) = \Delta T_0 e^{-kt}$.We want to find the time $t$ when $\Delta T(t) = \frac{\Delta T_0}{2}$.Substituting this into the equation: $\frac{\Delta T_0}{2} = \Delta T_0 e^{-kt}$.This simplifies to $\frac{1}{2} = e^{-kt}$,or $2 = e^{kt}$.Taking the natural logarithm on both sides: $\ln(2) = kt$.Therefore,$t = \frac{\ln(2)}{k}$.Substituting $k = \frac{R}{\Delta T_0}$,we get $t = \frac{\ln(2) \Delta T_0}{R}$.

For a system where Newton's law of cooling is applicable,the initial rate of cooling is $R \ ^\circ C/sec$. Find the time when the temperature difference $\Delta T_0$ (initial temperature difference) is reduced to half.

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