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Emissive and Absorptive Power

Two electric ...

Question

Two electric lamps $A$ and $B$ radiate the same power. Their filaments have the same dimensions, and have emissivities $e_{A}$ and $e_{B}$ . Their surface temperatures are $T_{A}$ and $T_{B}$ . The ratio $\frac{T_{A}}{T_{B}}$ will be equal to

{(\frac{e_{B}}{e_{A}})}^{1 / 4}

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{(\frac{e_{B}}{e_{A}})}^{1 / 2}

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{(\frac{e_{A}}{e_{B}})}^{1 / 2}

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{(\frac{e_{A}}{e_{B}})}^{1 / 4}

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Solution

The correct option is A ${(\frac{e_{B}}{e_{A}})}^{1 / 4}$
$P o w e r r a d i a t e d d u e t o r a d i a t i o n i s, P = e A σ T^{4}, w h e r e e = e m m i s i v i t y σ = S t e f a n^{'} s c o n s t a n t A = S u r f a c e a r e a o f t h e b o d y N o w, i n t h e q u e s t i o n, s i n c e P, A a n d σ a r e c o n s t a n t s e T^{4} = c o n s t a n t \Rightarrow e_{A} {T_{A}}^{4} = e_{B} {T_{B}}^{4} \Rightarrow \frac{T_{A}}{T_{B}} = {(\frac{e_{B}}{e_{A}})}^{\frac{1}{4}}$

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Two electric lamps A and B radiate the same power. Their filaments have the same dimensions, and have emissivities eA and eB. Their surface temperatures are TA and TB. The ratio TATB will be equal to

The correct option is A (eBeA)1/4Powerradiatedduetoradiationis,P=eAσT4,wheree= emmisivityσ= Stefan′s constantA= SurfaceareaofthebodyNow,inthequestion,sinceP,AandσareconstantseT4=constant⇒eATA4=eBTB4⇒TATB=(eBeA)14

Two electric lamps $A$ and $B$ radiate the same power. Their filaments have the same dimensions, and have emissivities $e_{A}$ and $e_{B}$ . Their surface temperatures are $T_{A}$ and $T_{B}$ . The ratio $\frac{T_{A}}{T_{B}}$ will be equal to