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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

Determine the amount of heat radiated per second by a body having a...

Determine the amount of heat radiated per second by a body having a surface area of 12 cm212 \text{ cm}^2, which is maintained in thermal equilibrium inside a room at a temperature of 20^{}^{{}^{\circ}} C. The surface emissivity is 0.800.80 and σ=6.0×10−8 W m−2 K−4\sigma = 6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}.

Options

  1. A0.53 W0.53 \text{ W}
  2. B0.42 W0.42 \text{ W}
  3. C4245 W4245 \text{ W}
  4. D0 W0 \text{ W}

Correct answer

B. 0.42 W0.42 \text{ W}

Step-by-step solution

The temperature of the body in thermal equilibrium with the room is T = 20^{}^{{}^{\circ}} \text{C} = 293 \text{ K}.

The surface area of the body is A=12 cm2=12×10−4 m2A = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2.

According to the Stefan-Boltzmann law, the amount of heat radiated per second by the body is given by:

P=eσAT4P = e \sigma A T^4

Substituting the given values:

P=0.80×(6.0×10−8)×(12×10−4)×(293)4P = 0.80 \times (6.0 \times 10^{-8}) \times (12 \times 10^{-4}) \times (293)^4

P=57.6×10−12×7.37×109P = 57.6 \times 10^{-12} \times 7.37 \times 10^9

P≈0.4245 WP \approx 0.4245 \text{ W}

Rounding off, the heat radiated per second is 0.42 W0.42 \text{ W}.
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