A body cools from `70^(@)C` to `50^(@)C` in 5minutes Temperature of surroundings is `20^(@)C` Its temperature after next `10 minutes is .

Options

A. `25^(@)C`
B. `30^(@)C`
C. `35^(@)`
D. `45^(@)C`

Answer

B

Explanation

According to Newton's law of cooling `(theta_(1)-theta_(2))/(t)=alpha[(theta_(1)+theta_(2))/(2)-theta_(0)]` for the given conditions `(70-50)/(5)=alpha[(70+50)/(2)-20].....(i) ` Let `theta` be the temperature after next 10 min. Then , `(50-theta)/(10)=alpha[(50-theta)/(2)-20].....(ii)`
Solving Eqs. (i) and (ii) , we get `theta=30^(@)C`.