Question

The graph shown in the adjacent diagram represents the variation of temperature $\left(\right.T\left.\right)$ of two bodies $x$ and $y$ having the same surface area, with time ( $t$ ). Both of these bodies lose heat only due to the emission of radiation. Find the correct relation between the emissive and absorptive power of the two bodies.

• A. $E_{x}>E_{y} \, and \, a_{x} < a_{y}$
• B. $E_{x} < E_{y} \, and \, a_{x}>a_{y}$
• C. $E_{x}>E_{y} \, and \, a_{x}>a_{y}$
• D. $E_{x} < E_{y} \, and \, a_{x} < a_{y}$

Answer 1

Answer: (C)

Rate of cooling $\left(- \frac{dT}{dt}\right) \propto $ emissivity $\left(e\right)$
From the graph,
$\left(- \frac{dT}{dt}\right)_{x}>\left(- \frac{dT}{dt}\right)_{y}$
$\therefore $ $e_{x}>e_{y}$
Further emissivity ( $e$ ) $ \propto $ absorptive power ( $a$ ) (good absorbers are good emitters also)
$\therefore $ $a_{x}>a_{y}$