A body cools from `50^(@)C` to `40^(@)C` in 5 minutes and from `40^(@)C` to `30^(@)C` in 8 minutes. Then the temperature of the surrounding is

Options

A. `25^(@)C`
B. `18^(@)C`
C. `15^(@)C`
D. `10^(@)C`

Answer

B

Explanation

Here `T_(1)=0^(@)C,T_(2)+40^(@)C,t+5` min Let `T_(c)` be temp. of curroundings
`:.` average temperature `T+(T_(1)+T_(2))/(2^(@)C)`
Change in temperature `dT+=0^(@)C`
`:.(dT)/(dt)+_k(T-T_(c))rArr(10)/(5)=-k(45-T_(c))` . . . (i) For `2^(nd)` case
`(10)/(8)=-k(35-T_(c))` . . . (ii)
Dividing (i) by (ii)
`(8)/(5)=(45-T_(c))/(35-T_(c))rArr8xx35-8T_(c)=5xx45-5T_(c)`
`T_(c)~=18^(@)C`
So correct choice is (b).