A glass of hot water cools from 90^ C to 70^ C in 3 minutes when the temperature of surroundings is 20^ C . What is the time taken by the glass of hot water to cool from 60^ C to 40^ C if the surrounding temperature remains the same at 20^ C ?

Step-by-step solution

According to Newton's law of cooling, the rate of cooling is given by dT dt = -k(T - T_s) , where T is the temperature of the body, T_s is the temperature of the surroundings, and k is a constant. For the first interval, T₁ = 90^ C , T₂ = 70^ C , T_s = 20^ C , and t = 3 minutes . Using the average temperature approximation T₁ - T₂ t = k ( T₁ + T₂ 2 - T_s ) , we get 90 - 70 3 = k ( 90 + 70 2 - 20 ) . 20 3 = k(80 - 20) = 60k , which implies k = 20 3 60 = 1 9 . For the second interval, T₁ = 60^ C , T₂ = 40^ C , T_s =