A metal sphere cools at a rate of 2^ C/min when the temperature is 80^ C . If the temperature of the surroundings is 30^ C , then the rate of cooling when its temperature is 50^ C will be

Step-by-step solution

According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the object and its surroundings. dT dt = k(T - T_s) Given that the rate of cooling is 2^ C/min when T = 80^ C and T_s = 30^ C 2 = k(80 - 30) 2 = 50k k = 2 50 = 1 25 When the temperature of the sphere is 50^ C , the new rate of cooling is dT dt = k(50 - 30) dT dt = 1 25 20 = 4 5 = 0.8^ C/min Answer: 0.8^ C/min