A body cools down from 52.5^(@)C to 47.5^(@)C in 5 min and to 42.5^(@)C in 7.5 min. Find the temperature of the surroundings.

A body cools down from 52.5^(@)C to 47.5^(@)C in 5 min and to 42.5^(@)

1.	A body cools down from 52.5^(@)C to 47.5^(@)C in 5 min and to 42.5^(@)C in 7.5 min. Find the temperature of the surroundings.
Answer» Solution :Using NEWTON's law of cooling. `(T_(1)-T_(2))/t=k((T_(1)-T_(1))/2-T_(0))` we get `(52.5^(@)C-42.5^(@)C)/(5min)= k((52.5^(@)C+47.5^(@)C)/(5min)-T_(0))` `(5^(@)`C)/(5min) = k(50^(@)C-T_(0))`............(i) `(47.5^(@)C-42.5^(@)C)/(7.5min)= k(47.5^(@)C+42.5^(@)C)/(2)-T_(0)` `(5^(@)C)/(7.5min)=k(45^(@)C-T_(0))` FORM (i) and (ii), we get `(50^(@)C-T_(0))/(45^(@)C-T_(0))= (7.5)/(5)=3/2` or `T_(0)=35^(@)`C