The surface area of a black body is 5 × 10^-4 ~m^2 and its temperature is 727^° C . The energy radiated by it per minute is (σ=5.67 × 10^-8 ~J / m^2-s-K^4 )

Correct answer: (A) 1.7 × 10^3 ~J. From Stefan's law E=σ T^4 A Given, T=727^° C=727+273=1000 ~K A=5 × 10^-4 ~m^2 ∴ Energy =5.67 × 10^-8 (1000)^4 (5 × 10^-4 ) 60 ∴ E=1.7 × 10^3 ~J

Q128 / 170 · Manipal (MET) 2017 · Physics

The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^2$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{m}^2-\mathrm{s}-\mathrm{K}^4\right)$

Correct answer
A
$1.7 \times 10^3 \mathrm{~J}$
B
$2.5 \times 10^2 \mathrm{~J}$
C
$8 \times 10^3 \mathrm{~J}$
D
$3 \times 10^4 \mathrm{~J}$

Solution

From Stefan's law $E=\sigma T^4 A$ Given, $T=727^{\circ} \mathrm{C}=727+273=1000 \mathrm{~K}$ $A=5 \times 10^{-4} \mathrm{~m}^2$ $\therefore$ Energy $=5.67 \times 10^{-8} \quad(1000)^4\left(5 \times 10^{-4}\right) 60$ $\therefore E=1.7 \times 10^3 \mathrm{~J}$

Hence, the correct answer is option: 'A'.

The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^2$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{m}^2-\mathrm{s}-\mathrm{K}^4\right)$

Solution

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