A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding.

Sol. 50°C 45°C 40°C Let the surrounding temperature be ‘T’°C Avg. t = 50 + 45/2 = 47.5 Avg. temp. diff. from surrounding T = 47.5 – T Rate of fall of temp = 50 – 45/5 = 1 °C/mm From Network’s Law 1°C/mm = bA * t ⇒ bA = 1/t = 1/47.5 – T …(1) In second case, Avg, temp = 40 + 45/2 = 42.5 Avg. temp. diff. from surrounding t’ = 42.5 – t rate of fall of temp = 45 – 40/8 = 5/8 °C/mm From Network’s Law ⇒ 5/B = 1/(47.5 - T) * (42.5 - T) By C & D [Componendo & Dividendo method] We find, T = 34.1°C