A liquid having a heat capacity of 1800 J K ⁻¹ cools from 80^ C to 60^ C in 12 minutes. The temperature of the surroundings is 20^ C . An electrical heater is now immersed in the liquid to maintain its temperature at a steady 80^ C . Assuming Newton's law of cooling is valid, the power of the heater required is ________ W .

Step-by-step solution

Using the average temperature approximation of Newton's law of cooling for the first process: T t = K ( T₁ + T₂ 2 - T₀ ) The liquid cools from 80^ C to 60^ C in 12 minutes: 80 - 60 12 = K ( 80 + 60 2 - 20 ) 20 12 = K(70 - 20) 5 3 = K(50) K = 1 30 min ⁻¹ To maintain the liquid at a steady temperature of 80^ C , the heater must supply heat at the same rate the liquid loses heat to the surroundings at 80^ C . The rate of cooling at 80^ C is: dT dt = K(T - T₀) = 1 30 (80 - 20) = 60 30 = 2^ C min ⁻¹ The rate of heat los