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A black coloured solid sphere of radius R and mass M is inside a cavity with vacuum inside. The walls of the cavity are maintained at temperature T₀. The initial temperature of the sphere is 3T₀. If the specific heat of the material of the sphere varies as α T³ per unit mass with the temperature T of the sphere, then the time taken for the sphere to cool down to temperature 2T₀ will be (σ is the Stefan–Boltzmann constant)

Asked in JEE Main Online 2014 · Stefan-Boltzmann law

Answer: (3) (Mα)/(16π R²σ)ln ((16)/3)

Step-by-step solution

Idea: the sphere sits in a cavity at T₀, so the net loss is σ A(T⁴-T₀⁴) — the surroundings term must be kept, and the area is the full surface 4π R².

Heat balance for a black sphere (e=1): Mc(dT)/(dt)=-σ(4π R²)(T⁴-T₀⁴) with c=α T³.

Mα(T³ dT)/(T⁴-T₀⁴)=-4π R²σ dt. Put u=T⁴-T₀⁴, so T³ dT=(du)/4.

(Mα)/4∫_80T₀⁴^15T₀⁴(du)/u=-4π R²σ t, since T=3T₀ gives u=80T₀⁴ and T=2T₀ gives u=15T₀⁴.

t=(Mα)/(16π R²σ)ln ((80)/(15))=(Mα)/(16π R²σ)ln ((16)/3).

Why the other options are wrong

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