Question

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 (1) of the two bodies

• A. $e_{x} > e_{y}\& a_{x} < a_{y}$
• B. $e_{x} < e_{y}\& a_{x} > a_{y}$
• C. $e_{x} > e_{y}\& a_{x} > a_{y}$
• D. $e_{x} < e_{y}\& a_{x} < a_{y}$

Answer 1

Answer: (C)

$e_{x} > e_{y}\& a_{x} > a_{y}$

Answer 2

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 then 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 Kirchoff's law good emitter are also good absorber so $a_{x} > a_{y}.$