A body cools down from `50^(@)C` to `45^(@)C` in `5` minutes and to `40^(@)C` in another `8` minutes. Find the temperature of the surrounding.

Explanation

Let `theta_(0)` is temperature of surrounding
According to Newton's law of cooling
`(Delta theta)/(Delta t) = - k (bar theta - theta_(0))`
where `Delta theta = theta_(1) = theta_(2) bar theta = (theta_(1) + theta_(2))/(2)`
`(50 - 45)/(5) = - k ((50 + 45)/(2) - theta_(0))`
`1 = - k (47.5 - theta_(0))`
`(45 - 40)/(8) = - k ((45 + 40)/(2) - theta_(0))`
`(5)/(8) = - k (42.5 - theta_(0))`
(i) (ii)
`(1)/(5//8) = (47.5 - theta_(0))/(42.5 - theta_(0))`
`8 (42.5 - theta_(0)) = 5 (47.5 - theta_(0))`
`340 - 8 theta_(0) = 237.5 - 5 theta_(0)`
`3 theta_(0) = 102.5`
`theta_(0) = 34.2^(@)C`