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

According two Newton's law of cooling
`(T_(1)-T_(2))/(Deltat)=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))`
On solving `T_(0)= 45^(@)C`