An electric heater emits 1000 W of thermal radiation. The coil has a surface area of 0.020 $m^{2}$ . Assuming that the coil radiates like a blackbody, find its temperature. $σ = 6.00 \times 10^{- 8} W / m^{2} - K^{4}$ .

955 deg C

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955 deg F

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955 K

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865 K

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Solution

The correct option is C
955 K

Let the temperature of the coil be T. The coil will emit radiation at a rate A $σ T^{4}$ . Thus,

1000W = $(0.020 m^{2}) \times (6.0 \times 12^{- 8} W / m^{2} - K^{4}) \times T 64$

or, $T^{4} = \frac{1000}{0.020 \times 6.00 \times 10^{- 8}} K^{4}$

= $8.33 \times 10^{11} K^{4}$

or, $T = 955 K$ .

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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=6.00×10−8W/m2−K4.

The correct option is C 955 K Let the temperature of the coil be T. The coil will emit radiation at a rate Aσ T4. Thus, 1000W =(0.020 m2)×(6.0×12−8W/m2−K4)×T64 or, T4=10000.020×6.00×10−8K4 =8.33×1011K4 or, T=955 K.

An electric heater emits 1000 W of thermal radiation. The coil has a surface area of 0.020 $m^{2}$ . Assuming that the coil radiates like a blackbody, find its temperature. $σ = 6.00 \times 10^{- 8} W / m^{2} - K^{4}$ .