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. `5^(@)C`
C. `15^(@)C`
D. `20^(@)C`

Answer

A

Explanation

Ac cording to Newton's law of cooling ltbRgt `(theta_(1)-theta_(2))/(t) = -k[(theta_(1) + theta_(2))/(2) - theta_(0)]`
`:. (60-50)/(10xx60) = -k[(60+50)/(2) - theta_(0)]= -k[55 - theta_(0)]` …..(i)
and `(50-42)/(10xx60) = -k[(50+42)/(2) - theta_(0)] = -k[46 - theta_(0)]` .....(ii)
Dividing 9i) by (ii) and solving we get, `theta_(0) =10^(@)C`