A body cools in 10 min from 60^(@)C to 40^(@)C in 5 min and to 42.5^(@)C in 7.5 min. Find the temperature of the surroundings.

A body cools in 10 min from 60^(@)C to 40^(@)C in 5 min and to 42.5^(@

1.	A body cools in 10 min from 60^(@)C to 40^(@)C in 5 min and to 42.5^(@)C in 7.5 min. Find the temperature of the surroundings.
Answer» Solution :According to Newton's LAW of cooling. `((T_(1)-T_(2))/(t))= K[((T_(1)+T_(2))/2)-T_(0)]` For the given conditions, `(60-40)/(10) = k[(60-40)/(2)-10]`.....................(i) Let T be the temperature after next 10 min. Then, `(40-T)/(10)=K[(40+T)/(2)-10]`..............(ii) SOLVING Eqs. (i) and (ii), we get T=`28^(@)`C