TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

According to Stefan's law heat energy absorbed is given by
‌

dQ

dt

=εσA(T4−T04)....(i)
where, T= temperature of body
and T0= temperature of surroundings.
If m be the mass of body and c be the specific heat, then heat loss at temperature T is
‌

dQ

dt

=−mc‌

dT

dt

...(ii)
From Eqs. (i) and (ii), we get
−mc‌

dT

dt

=εσA(T4−T0−4)
⇒‌‌−‌

dT

dt

=‌

εσA

mc

(T4−T04)
Here, T=T0+∆T
−‌

dT

dt

=‌

εσA

mc

[(T0+∆T)4−T04]
=‌

εσA

mc

T04[(1+‌

∆T

T0

)4−1]
$=\;{ε σ A}/{m c} T_{0}^{4}[1+\;{4 ∆ T}/{T_{0}}-1]( ∵ ∆ T< =‌

εσA4T03

mc

∆T
⇒‌‌−‌

dT

dt

=k∆T
wherc, ‌‌k=‌

εσA4T03

mc

∴‌‌‌

dT

dt

=−k(T−T0)
This is Newton's law of cooling.
Hence, Newton's law of cooling is a special case of Stefan's law.