# 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.

Deepak Patra
10 years ago
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
4 years ago
Dear student,

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
bA = 5/8t’ = 5/8(42.5 - T)   …..(2)
from eqn (1) and (2)
⇒ 5/8 = (42.5 – T)/(47.5 – T)
By Componendo & Dividendo method
We find, T = 34.1°C

Thanks and regards,
Kushagra