A very thin metallic shell of radius r is hea | vedclass

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(D) According to Newton's Law of Cooling,the rate of cooling $(dT/dt)$ is given by the formula: $\frac{dT}{dt} = -k A (T - T_s)$,where $A$ is the surf

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$r^0$

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(D) According to Newton's Law of Cooling,the rate of cooling $(dT/dt)$ is given by the formula: $\frac{dT}{dt} = -k A (T - T_s)$,where $A$ is the surface area,$T$ is the temperature of the body,and $T_s$ is the surrounding temperature.For a spherical shell,the surface area $A = 4 \pi r^2$.The rate of cooling is also defined as the rate of loss of heat per unit heat capacity: $\frac{dT}{dt} = \frac{dQ/dt}{ms}$,where $m$ is the mass and $s$ is the specific heat capacity.Since $m = \rho V = \rho (4 \pi r^2 \delta)$,where $\delta$ is the thickness of the shell,we have:$\frac{dT}{dt} = \frac{\sigma A (T^4 - T_s^4)}{ms} \propto \frac{r^2}{r^2} \propto r^0$.Thus,the rate of cooling is independent of the radius $r$.

$A$ very thin metallic shell of radius $r$ is heated to temperature $T$ and then allowed to cool. The rate of cooling of the shell is proportional to ........

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