Dear student,
Please find the solution to your problem.
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
