Given: T1 = 273 + 727 = 1000 K, T2 = 273 + 227 = 500 K, T0 = 273 + 27 = 300 K
i) The rate of emission of heat,
dQ/dt = σAT4
we assume that the surface area A is the same for the two bodies.
\(\frac{(dQ/dt)_1}{(dQ/dt)_2}\) = \(\frac{T_1^4}{T_2^4}\)
= \((\frac{T_1}{T_2})^4\)
=\((\frac{1000}{500})^4\) = 24 = 16
i) The rate of loss of heat,
dQ'/dt = σA( T4 - T04 )
\(\frac{(dQ'/dt)_1}{(dQ/dt)_2}\) = \(\frac{T_1^4-T_0^4}{T_2^4-T_0^4}\)
= \(\frac{10^{12}-81×10^8}{625×10^8-81×10^8}\)
= 18.23