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The area of a hose of heat furnace is 10^{-4}m^2 . It radiates 1.58\times 10^5 cal of heat per hour. If the emissivity of the furnace is 0.80, then its temperature (in K) is

Option: 1 1500

Option: 2 2000

Option: 3 2500

Option: 4 3000

Answers (1)

best_answer

As we have learnt,

 

For Ordinary Body -

e=\epsilon E=A\epsilon \sigma \theta^{4}

- wherein

\epsilon = represent emissivity of the material

 

 E = \sigma\epsilon AT^4

or

\frac{1.58\times 10^5\times 4.2}{60\times 60} = 5.6\times 10^{-8}\times 10^{-4}\times0.8\times T^4

or 

T \approx 2500K

 

Posted by

Sanket Gandhi

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