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The wavelength-specific emissivity $\epsilon_{\lambda}$ of a body is the ratio of the body's spectral radiance at the specific wavelength compared to that of the ideal blackbody. Does $\epsilon_{\lambda}$ depend on temperature?

If so, it would mean that an object with uniform $\epsilon_{\lambda}$ across all wavelength at one temperature might no longer have this property at another temperature. So am I right to say that the property of being a "grey body" is temperature dependent?

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Yes, it depends on temperature : $$ \begin{align} \varepsilon _{\lambda }&={\frac {M_{\mathrm {e} ,\lambda }}{M_{\mathrm {e} ,\lambda }^{\circ }}} \\&= {{\frac {\lambda ^{5} \left({e^{\frac {\mathrm {h} c}{\lambda \mathrm {k} T}}-1}\right)}{2\pi \mathrm {h} c^{2}}}} \cdot {\frac {\partial M_{\mathrm {e} }}{\partial \lambda }} \\&={{\frac {\varepsilon \sigma \lambda ^{5} \left({e^{\frac {\mathrm {h} c}{\lambda \mathrm {k} T}}-1}\right)}{2\pi \mathrm {h} c^{2}}}} \cdot {\frac {\partial T^{4} }{\partial \lambda }} \end{align} $$ where $\varepsilon$ is the emissivity coefficient of material surface and $\sigma$ is the Stefan–Boltzmann constant. So for being able to calculate wavelength-specific emissivity of body, you need to know it's temperature and it's temperature distribution law with respect to emitted wavelength (derivative part in equation).

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