The ratio of emissive power of perfect blackbody at $1327^\circ$C and $527^\circ$C is ______.

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The ratio of emissive power of perfect blackbody at $1327^\circ$C and $527^\circ$C is 16:1. Explanation: The temperature of the black body is T’ = $1327^\circ$C. The temperature of the black body is T’’ = $527^\circ$C. Express the relation between emissive power and temperature. E_b = n^2sigma T^4 Here, E_b is emissive power, n is an index of refraction, sigma is Stefan’s constant, T is temperature. Express the relation of emissive power at temperature $1327^\circ$C. E'b = n^2 sigma T"’’"^4 .....(i) Here, E'b is emissive power at temperature $1327^\circ$C. Express the relation of emissive power at temperature $527^\circ$C E''b = n^2 sigma T"’"^4 .....(ii) Here, E′′b s emissive power at temperature $527^\circ$C Convert temperatures T' and T'' from degree Celsius to kelvin. T' = $1327^\circ$C + 273 T' = 1600 K T'' = $527^\circ$C + 273 T'' = 800 K Substitute 1600K for T' and 800 K for T'' in equation (i) and (ii) respectively to find emissive power. E'b = nsigma (1600)^4 E''b = nsigma (800)^4 Divide E'b by E''b to find the relation between E'b and E''b. `(E"'"_b)/(E"''"_b) = (nsigma(1600)^4)/(nsigma(800)^4 (E"'"_b)/(E"''"_b) = ($\frac{1600}{800}$)^4 (E"'"_b)/(E"''_b) = (2)^4 (E"'"_b)/(E"''"_b)= $\frac{16}{1}$`

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