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Standard VIII

Physics

Sources of Heat

The area of a...

Question

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

1500 K

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2000 K

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2500 K

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3000 K

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Solution

The correct option is C
2500 K

Acoording to stefan's law $E = σ ε A T^{4}$
$\Rightarrow \frac{1.58 \times 10^{5} \times 4.2}{60 \times 60} = 5.6 \times 10^{- 8} \times 10^{- 4} \times 0.8 \times T^{4}$
$T \approx 2500 K$

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Similar questions

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The area of a hole of heat furnace is 10−4m2. It radiates 1.58×105 calories of heat per hour. If the emissivity of the furnace is 0.80, then its temperature is

The correct option is C 2500 K Acoording to stefan's law E=σεAT4 ⇒1.58×105×4.260×60=5.6×10−8×10−4×0.8×T4 T≈2500 K

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

The correct option is C
2500 K

Acoording to stefan's law $E = σ ε A T^{4}$
$\Rightarrow \frac{1.58 \times 10^{5} \times 4.2}{60 \times 60} = 5.6 \times 10^{- 8} \times 10^{- 4} \times 0.8 \times T^{4}$
$T \approx 2500 K$