Two metal spheres,one solid and one hollow,ar | vedclass

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(C) According to Newton's Law of Cooling,the rate of heat loss $(dQ/dt)$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$,where $e$ is emissivity,$\sigm

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Equal for both

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(C) According to Newton's Law of Cooling,the rate of heat loss $(dQ/dt)$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$,where $e$ is emissivity,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,$T$ is the temperature of the body,and $T_0$ is the temperature of the surroundings.Since both spheres are made of the same metal and are heated to the same temperature $T$ and placed in the same environment $T_0$,the rate of heat loss depends on the surface area $A$ and emissivity $e$.If the spheres have the same radius,they have the same surface area $A = 4\pi r^2$.Assuming the spheres have the same radius,the rate of heat loss will be the same for both,provided the emissivity is also the same.

Two metal spheres,one solid and one hollow,are heated to $300^{\circ}C$ and allowed to cool in the same environment. The rate of heat loss will be:

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