JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 1 Questions

In the given problem, fall in temperature of sphere,
dT=(3T0−2T0)=T0
Temperature of surrounding, Tsurr =T0
Initial temperature of sphere, Tinitial =3T0
Specific heat of the material of the sphere varies as,
c=αT3 per unit mass (α= a constant )
Applying formula,

dT

dt

=

σA

McJ

(T4−Tsurr4)
⇒

T0

dt

=

σ4πR2

Mα(3T0)3J

[(3T0)4−(T0)4]
⇒dt=

Mα27T04J

σ4πR2×80T04

Solving we get,
Time taken for the sphere to cool down temperature 2T0,
T=

Mα

16πR2σ

In(

16

3

)