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Convection

Assuming Newt...

Question

Assuming Newton's law of cooling to be valid, the time taken by a body to cool from a temperature $T_{1}$ to a temperature $T_{2}$ in a room where the temperature is $T_{0}$ $(T_{0} < T_{2} < T_{1})$ is proportional to

$l o g_{e} (\frac{T_{1} - T_{0}}{T_{2} - T_{0}})$

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$l o g_{e} (\frac{T_{2} - T_{0}}{T_{1} + T_{0}})$

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$l o g_{e} (\frac{T_{1} - T_{2}}{T_{0}})$

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$l o g_{e} (\frac{T_{1} + T_{0}}{T_{2} + T_{0}})$

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Solution

The correct option is A
$l o g_{e} (\frac{T_{1} - T_{0}}{T_{2} - T_{0}})$

According to Newton's law of cooling, the rate of fall of temperature is given by
$\frac{d T}{d t} = - K (T - T_{0})$
where K is a constant, T is the temperature of the body at time t and $T_{0}$ that of its surrounding.
Hence, $- K d t = \frac{d T}{T - T_{0}}$
Integrating, we have
$- K \int_{0}^{t} d t = \int_{T_{1}}^{T_{2}} \frac{d T}{T - T_{0}} \Rightarrow - K t = | l o g_{e} (T - T_{0}) |_{T_{1}}^{T_{2}} = l o g_{e} (\frac{T_{2} - T_{0}}{T_{1} - T_{0}}) = - l o g_{e} (\frac{T_{1} - T_{0}}{T_{2} - T_{0}}) \Rightarrow t = \frac{1}{K} l o g_{e} (\frac{T_{1} - T_{0}}{T_{2} - T_{0}})$
Hence, the correct choice is (a).

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Assuming Newton's law of cooling to be valid, the time taken by a body to cool from a temperature T1 to a temperature T2 in a room where the temperature is T0 (T0<T2<T1) is proportional to

Assuming Newton's law of cooling to be valid, the time taken by a body to cool from a temperature $T_{1}$ to a temperature $T_{2}$ in a room where the temperature is $T_{0}$ $(T_{0} < T_{2} < T_{1})$ is proportional to