Hot water cools from `60^@C` to `50^@C` in the first 10 min and to `42^@C` in the next 10 min. The temperature of the surrounding is

Options

A. `10^(@)C`
B. `15^(@)C`
C. `20^(@)C`
D. `30^(@)C`

Answer

A

Explanation

As per Newton's law of cooling,
`T_(1)-T_(2)=K[(T_(1)+T_(2))/(2)-T_(s)]`
where , `T_(s)` is temperature of surrounding .
`(60-50)/(10)=K[(60+50)/(2)-T_(s)]`
`1=K[55-T_(s)]`
`"Similarly "(50-42)/(10)=K(46-T_(s)),`
`(8)/(10)=K(46-T_(s))` ...(i)
Dividing Eqs . (i) and (ii) , we have `(10)/(8)=(k(55-T_(s)))/(k(46-T_(s)))`
`rArr" " T_(s)=10^(@)C`