Certain quantity of water cools from `70^(@)C` to `60^(@)C` in the first `5` minutes and to `54^(@)C` in the next `5` minutes. The temperature of the surrounding is

Options

A. `45^(@)C`
B. `20^(@)C`
C. `42^(@)C`
D. `10^(@)C`

Answer

A

Explanation

Ac cording to Newton's law of cooling
`(T_(1)-T_(2))/(Delta t) =-K [(T_(1)+T_(2))/(2)-T_(0)]`
Case (i) `(70-60)/(5)= -K [(70+60)/(2)-T_(0)]`
or `2= -K[65 - T_(0)]` …..(i)
Case (ii) `(60-54)/(5)= -K [(60+54)/(2) - T_(0)]`
or `(6)/(5)= -K[57 - T_(0)]` ....(ii)
Dividing (i) by (ii), we get
`(5)/(3) =(65 - T_(0))/(57 - T_(0))1`
On solving, `T_(0) =45^(@)C` .