If transmittance of a surface is \(\frac{1}{7}\), reflectance is \(\frac{1}{8}\), then the absorptance of the surface will be
By conservation of energy, the sum of absorptance \(a\), reflectance \(r\), and transmittance \(t\) is equal to \(1\): \(a + r + t = 1\). Therefore, \(a = 1 - \frac{1}{8} - \frac{1}{7} = 1 - \frac{15}{56} = \frac{41}{56}\).
The unit of emissive power is
Emissive power is defined as the thermal energy radiated per unit area per unit time. Its SI unit is \(\text{J m}^{-2}\text{s}^{-1}\) (or \(\text{W m}^{-2}\)).
A blackbody and a real body of identical dimensions are heated to same temperature. If ratio of rates of radiation of the blackbody and the real body is $4 : 3$, then emissivity of the real body is equal to
The rate of radiation is given by $E = e\sigma A T^4$. For a blackbody, $e = 1$. Given $\frac{E_b}{E} = \frac{4}{3} ⇒ \frac{1}{e} = \frac{4}{3}$, hence emissivity $e = 0.75$.
The unit of emissive power is
Emissive power is defined as the thermal energy emitted per unit area per unit time, so its unit is \(\text{J m}^{-2}\text{ s}^{-1}\) (or \(\text{W m}^{-2}\)).