Practice portal › Thermal Properties of Matter › Radiation, Stefan's Law and Cooling
Asked in JEE Main Online 2014 · Stefan-Boltzmann law
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).
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