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A hot body placed in air is cooled down ...

A hot body placed in air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being `k` times the temperature difference from the surrounding. Starting from `t=0` , find the time in which the body will lose half the maximum heat it can lose.

Text Solution

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We have,
`-(dT)/(dt)k(T-T_0)`
where `T_0` is the temperature of the surrounding. If `T_1` is the initial temperature and `T` is the temperature at any time `t`, then
`int_(T_1)^(T)(dT)/((T-T_0))=-kint_0^t`
or `|ln(T-T_0)|_T^T=-kt` or `ln[(T-T_0)/(T_1-T_0)]=-kt`
or `T=T_0+(T_1-T_0)e^(-kt)` ..(i)
The body continues to lose heat till its temperature becomes equal to that of the surrounding. The loss of heat
`Q=mc(T_1-T_0)`
If the body half of the maximum loss that it can, then decrease in temperature
`(Q)/(2)=mc((T_1-T_0)/(2))`
If body loses this heat in time t, then its temperature at time t' will be
`T_1-((T_1-T_0)/(2))=(T_1+T_0)/(2)`
Putting these values in Eq. (i), we have
`(T_1-T_0)/(2)=T_0+(T_1-T_0)e^(-kt)`
or `(T_1-T_0)/(2)=(T_1-T_0)e^(-kt)` or `e^(-kt)(1)/(2)` or `t'=(ln 2)/(k)`.
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