A hot fbody 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 loss half the maximum heat it can lose.

Explanation

We have
`(d theta )/(dt)=-k(theta-theta_(0)` . Where `theta_(0` . Is the temperature of the surrounding and theta his the temperature of the body at time t. Suppose `theta=theta_(1)` at t=0 .
Then, ` int_(theta1)^(theta)(d theta)/(theta-theta_(0))=-kint_(0)^(t)dt` .or, `ln`(theta-theta_(0))/(theta_(1)-theta_(0))=-kt` . or, `theta-theta_(0)=(theta_(1)-theta_(0))e-^(kt)` . The body continues to lose heat till its temperature become equal to that of the surrounding. The loss of heat in this entier periodis `DeltaQ_(m)=ms(theta_(1)-theta_(0))` . This is the maximum heat the body can lose. If the body loses half this heat, the decrreases in its temoerature will be,
`(DeltaQ_(m))/(2ms)=(theta_(1)-theta_(0))/(2)` . If the body losses this heat in time t_(1,) the temperature at t_(1)` . will be
`theta_(1)-(theta_(1)-theta_(0))/(2)=(theta_(1)-theta_(0))/(2)` . Putting these values of time and temperature in (i),
`(theta_(1)+theta_(0))/(2)-theta_(0)=(theta_(1)-theta_(0))e^(-kt_(1))` . or, `e^(-kt_(1))=1/2` . or, `t_(1)=(ln2)/(k)` .