The excess temperature of a hot body above its surroundings is halved in t = 10 minutes. In what time will it be `(1)/(10)` of its initial value. Assume Newton's law of cooling.

Explanation

Using Newton's law of cooling we have
`(T-T_(s)) = (T_(0) -T_(s))e^(-kt)`
at `t =10 min(T-T_(s)) = (T_(0)-T_(s))/(2)`
`rArr e^(-k(10)) =(1)/(2)`
or `k = (In(2))/(10)` Then at `t = t_(0)`
`T-T_(s) =(T_(0)-T_(s))/(10)`
`rArr e^(-kt)0 = (1)/(10)`
`t_(0) (In(10))/(In(2)) xx 10`
=33.23 min