An object is at a temperature of $400$ $^{o}$ C. The temperature at which it can radiate energy twice as faster (Neglect the temperature of surroundings) is ( $2^{\frac{1}{4}} =$ 1.189) :

800 K

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673 K

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500 K

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400 K

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Solution

The correct option is A 800 K
From Stefan-boltzman law $E_{b} = 5.67 \times {(\frac{T}{100})}^{4}$
So applying $E_{1} = {2 E}_{2}$
$5.67 \times {(\frac{T}{100})}^{4} = 2 \times 5.67 \times {(\frac{673}{100})}^{4}$

$T = 800.33 K$

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An object is at a temperature of 400oC. The temperature at which it can radiate energy twice as faster (Neglect the temperature of surroundings) is ( 214= 1.189) :

The correct option is A 800 KFrom Stefan-boltzman law Eb=5.67×(T100)4So applying E1=2E25.67×(T100)4=2×5.67×(673100)4T=800.33K

An object is at a temperature of $400$ $^{o}$ C. The temperature at which it can radiate energy twice as faster (Neglect the temperature of surroundings) is ( $2^{\frac{1}{4}} =$ 1.189) :

The correct option is A 800 K
From Stefan-boltzman law $E_{b} = 5.67 \times {(\frac{T}{100})}^{4}$
So applying $E_{1} = {2 E}_{2}$
$5.67 \times {(\frac{T}{100})}^{4} = 2 \times 5.67 \times {(\frac{673}{100})}^{4}$

$T = 800.33 K$