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Question
The ratio of emissive power of perfect blackbody at 1327°C and 527°C is ______.
Options
4:1
16:1
2:1
8:1
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Solution
The ratio of emissive power of perfect blackbody at 1327°C and 527°C is 16:1.
Explanation:
The temperature of the black body is T’ = 1327°C.
The temperature of the black body is T’’ = 527°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°C.
E'b = `n^2 sigma T"’’"^4` .....(i)
Here, E'b is emissive power at temperature 1327°C.
Express the relation of emissive power at temperature 527°C
E''b = `n^2 sigma T"’"^4` .....(ii)
Here, E′′b s emissive power at temperature 527°C
Convert temperatures T' and T'' from degree Celsius to kelvin.
T' = 1327°C + 273
T' = 1600 K
T'' = 527°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) = (1600/800)^4`
`(E"'"_b)/(E"''_b) = (2)^4`
`(E"'"_b)/(E"''"_b)= 16/1`
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