A solid cube and a solid sphere of identical | vedclass

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(B) According to Newton's law of cooling,the rate of heat loss $dQ/dt$ is proportional to the surface area $A$ of the body,i.e.,$dQ/dt \propto A$.For

Vedclass · Accepted Answer

the cube will cool faster because it has a larger surface area

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(B) According to Newton's law of cooling,the rate of heat loss $dQ/dt$ is proportional to the surface area $A$ of the body,i.e.,$dQ/dt \propto A$.For a given mass $m$ and density $\rho$,the volume $V = m/\rho$ is the same for both the cube and the sphere.Since the volume of a sphere is $V = (4/3)\pi r^3$ and the volume of a cube is $V = a^3$,the surface area of the cube $(6a^2)$ is greater than the surface area of the sphere $(4\pi r^2)$ for the same volume.Because the cube has a larger surface area,it loses heat at a faster rate than the sphere.Additionally,sharp edges on the cube facilitate more effective radiation compared to a smooth spherical surface.Therefore,the cube cools faster than the sphere. Thus,option $(b)$ is correct.

$A$ solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,

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