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Question: The area of a hole of heat furnace is \(10 ^ { - 4 } m ^ { 2 }\). It radiates \(1.58 \times 10 ^ { 5...

The area of a hole of heat furnace is 10−4m210 ^ { - 4 } m ^ { 2 }. It radiates 1.58×1051.58 \times 10 ^ { 5 }calories of heat per hour. If the emissivity of the furnace is 0.80, then its temperature is

A

1500 K

B

2000 K

C

2500 K

D

3000 K

Answer

2500 K

Explanation

Solution

According to Stefen’s law

E=σεAT4E = \sigma \varepsilon A T ^ { 4 }

⇒1.58×105×4.260×60=5.6×10−8×10−4×0.8×T4\frac { 1.58 \times 10 ^ { 5 } \times 4.2 } { 60 \times 60 } = 5.6 \times 10 ^ { - 8 } \times 10 ^ { - 4 } \times 0.8 \times T ^ { 4 }

T≈2500 KT \approx 2500 \mathrm {~K}