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TS EAMCET · Physics · Thermal Properties of Matter

The surface of a black body is at a temperature \(727^{\circ} \mathrm{C}\) and its cross-section is \(1 \mathrm{~m}^2\). Heat radiated from this surface in one minute in joules is (Stefan's constant \(=5.7 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 / \mathrm{k}^4\) )

  1. A \(34.2 \times 10^5\)
  2. B \(2.5 \times 10^5\)
  3. C \(3.42 \times 10^5\)
  4. D \(2.5 \times 10^6\)
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Answer & Solution

Correct Answer

(A) \(34.2 \times 10^5\)

Step-by-step Solution

Detailed explanation

Given, Temperature of black body \((T)=727^{\circ} \mathrm{C}\) \[ =727+273=1000 \mathrm{~K} \] Cross-sectional area of black body \((A)=1 \mathrm{~m}^2\) Stefan's constant \((\sigma)=5.7 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 / \mathrm{K}^4\) We know that, Heat radiated per…
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