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 ($a$) power of the two bodies. (IIT JEE 2003)

• A. e_x>e_y, a_x> a_y
• B. e_x>e_y, a_y> a_x
• C. e_y>e_x, a_y> a_x
• D. e_y>e_x, a_x> a_y

Answer 1

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 $\therefore e_x>e_y$ and according to Kirchhoff's law good emitter are also good absorber so $a_x>a_y$.