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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2012 · 7 May · Online · Physics · Thermal Properties of Matter

The heat radiated per unit area in 1 hour by a furnace whose...

The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K3000 \mathrm{~K} is (σ=5.7×(\sigma=5.7 \times 10−8 W m−2 K−410^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4} )

Options

  1. A1.7×1010 J1.7 \times 10^{10} \mathrm{~J}
  2. B1.1×1012 J1.1 \times 10^{12} \mathrm{~J}
  3. C2.8×108 J2.8 \times 10^8 \mathrm{~J}
  4. D4.6×106 J4.6 \times 10^6 \mathrm{~J}

Correct answer

A. 1.7×1010 J1.7 \times 10^{10} \mathrm{~J}

Step-by-step solution

According to Stefan's law E=σ\sigmaT^4 Heat radiated per unit area in 1 hour ( 3600 s)3600 \mathrm{~s}) is =5.7 ×\times10^{-8} ×(\times(300)^4 ×\times3600=1.7 ×\times10^{10}  J\mathrm{~J}
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