A body cools from 80^ C to 50^ C in 5 min . In the next time of ' t ' min, the body continues to cool from 50^ C to 30^ C . The total time taken by the body to cool from 80^ C to 30^ C is [The temperature of the surroundings is 20^ C .]

Step-by-step solution

The temperature of the surroundings is T_s = 20^ C . Newton's Law of Cooling in approximate finite difference form states: T₁ - T₂ t = K ( T₁ + T₂ 2 - T_s ) First, determine the cooling constant K using the cooling from 80^ C to 50^ C in 5 , min : 80 - 50 5 = K ( 80 + 50 2 - 20 ) 6 = K (45) K = 2 15 , min ⁻¹ Now, apply this K to find the time t to cool from 50^ C to 30^ C : 50 - 30 t = 2 15 ( 50 + 30 2 - 20 ) 20 t = 2 15 (20) t = 7.5 , min The total time to cool from 80^ C to 30^ C is 5 + 7.5 = 12.5 , min .