A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet, hung in the open air, loses half its moisture during the first hour, then 90 % of the moisture will be lost in ...... hours.

Step-by-step solution

The moisture loss rate is proportional to the current moisture content, leading to the differential equation dM dt = -kM with k > 0 . This separable equation integrates to |M| = -kt + C , yielding the general solution M(t) = M₀ e^ -kt where M₀ is the initial moisture. Using the condition that half the moisture is lost in the first hour, M(1) = 1 2 M₀ , substitute to find 1 2 = e^ -k , so k = 2 . The solution becomes M(t) = M₀ e^ -( 2)t = M₀ 2^ -t . To find when 90% is lost, set M(t) = 0.10 M₀ , giving 1 10 = 2^ -t