A solid cube and a solid sphere of the same m | vedclass

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Question

Rate of cooling of a body

$\mathrm{R}=\frac{\Delta 	heta}{\mathrm{t}}=\frac{\mathrm{A} \varepsilon \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}

Vedclass · Accepted Answer

The cube cools down faster than the sphere

Vedclass · Answer

Rate of cooling of a body

$\mathrm{R}=\frac{\Delta 	heta}{\mathrm{t}}=\frac{\mathrm{A} \varepsilon \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}ight)}{\mathrm{mc}}$

$\Rightarrow \mathrm{R} \propto \frac{\mathrm{A}}{\mathrm{m}} \propto \frac{	ext { Area }}{	ext { Volume }}$

$\Rightarrow$ For the same surface area. $\mathrm{R} \propto \frac{1}{	ext { Volume }}$

$\because$ Volume of cube $<$ volume of sphere

$\Rightarrow \mathrm{R}_{	ext {cute }}>\mathrm{R}_{	ext {Sphere }}$ i.e., cube, cools down with faster rate

A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120\ ^oC$ , then

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