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Question
An electric heater emits 1000 W of thermal radiation. The coil has a surface area of 0.020 m2. Assuming that the coil radiates like a blackbody, find its temperature. σ=6.00×108W/m2K4.
Options:
A .  955 deg C
B .  955 deg F
C .  955 K
D .  865 K
Answer: Option C
:
C
Let the temperature of the coil be T. The coil will emit radiation at a rate AσT4. Thus,
1000W =(0.020m2)×(6.0×128W/m2K4)×T64
or, T4=10000.020×6.00×108K4
=8.33×1011K4
or, T=955K.

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