The graph shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity ( $e$ ) and absorptivity ( $a$ ) power of the two bodies. (IIT JEE 2003)

$e_{x} > e_{y}, a_{x} > a_{y}$

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$e_{x} > e_{y}, a_{y} > a_{x}$

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$e_{y} > e_{x}, a_{y} > a_{x}$

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$e_{y} > e_{x}, a_{x} > a_{y}$

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Solution

The correct option is A
$e_{x} > e_{y}, a_{x} > a_{y}$

From the graph it is clear that initially both the bodies are at same temperature but after that at any instant temperature of body x is less than the temperature of body y. It means body x emits more heat i.e., emissivity of body x is more than body y $∴ e_{x} > e_{y}$ and according to Kirchhoff's law good emitter are also good absorber so $a_{x} > a_{y}$ .

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The graph shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity ( $e$ ) and absorptivity ( $a$ ) power of the two bodies. (IIT JEE 2003)