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Question

Physics Question on thermal properties of matter

The temperature of a body is increased from 73C-73^{\circ}\,C to 327C327\,^{\circ}\,C. Then the ratio of emissive power is

A

44570

B

44588

C

27-Jan

D

Jan-81

Answer

Jan-81

Explanation

Solution

Radiation from a body = R
R = e σT4\sigma T^{4} ? time ? area, where e = emissive power
? eT4\quad eT^{4} = constant or e =(constantT4)\left(\frac{constant}{T^{4}}\right)
or e1e2=(T2T1)4\frac{e_{1}}{e_{2}}=\left(\frac{T_{2}}{T_{1}}\right)^{^4} or e1e2=(273+32727373)4\frac{e_{1}}{e_{2}}=\left(\frac{273+327}{273-73}\right)^{^4}
=(600200)4=(3)41=811=\left(\frac{600}{200}\right)^{^4}=\frac{\left(3\right)^{4}}{1}=\frac{81}{1}
or e1e2=811\frac{e_{1}}{e_{2}}=\frac{81}{1}