Question

A body cools from `50^(@)C` to `49.9^(@)C` in `5` sec. it cools from `40^(@)C` to `39.9^(@)C` in t sec. Assuming Newtons law of cooling to be valid and temperature of surrounds at `30^(@)C`, value of t//5 will be?

Answer 1

Correct Answer - 2
From `(theta_(1)-theta_(2))/(t) = k (theta_(1)+theta_(2)/(2)- theta_(0))`
We have `(0.1)/(5) = "k"(39.9)/(2)` and `(0.1)/(t) =k(19.9)/(2) implies (t)/(5) = (39.9)/(19.9) = 2 implies t = 10 implies (t)/(5) = 2`