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(B) When a body cools by radiation,the rate of cooling $R$ is given by Newton's Law of Cooling as:$R = -\frac{d	heta}{dt} = \frac{eA\sigma}{ms}(	het

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Specific heat of $A$ is less

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(B) When a body cools by radiation,the rate of cooling $R$ is given by Newton's Law of Cooling as:$R = -\frac{d	heta}{dt} = \frac{eA\sigma}{ms}(	heta^4 - 	heta_0^4)$For small temperature differences $(	heta - 	heta_0)$,this simplifies to:$R \approx \frac{4eA\sigma	heta_0^3}{ms}(	heta - 	heta_0)$Since the discs have equal radii and are blackened,their surface area $A$ and emissivity $e$ are the same. Given that they are cooled under identical conditions,the surrounding temperature $	heta_0$ is also the same. Thus,the rate of cooling $R$ is inversely proportional to the mass $m$ and specific heat $s$ of the material:$R \propto \frac{1}{ms}$Assuming the discs have the same mass,$R \propto \frac{1}{s}$.From the given graph,the slope of the line represents the rate of cooling for a given temperature difference. The slope of line $A$ is greater than the slope of line $B$,which means the rate of cooling for disc $A$ is higher than that for disc $B$.Since $R_A > R_B$,it follows that $s_A < s_B$. Therefore,the specific heat of $A$ is less than the specific heat of $B$.

Two circular discs $A$ and $B$ with equal radii are blackened. They are heated to the same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves shown in the figure?

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