TS EAMCET · Physics · Thermal Properties of Matter
Heat loss takes place from a body maintained at a temperature of \(400^{\circ} \mathrm{C}\) to the surrounding air at \(30^{\circ} \mathrm{C}\) by convection and to the surrounding surfaces at \(30^{\circ} \mathrm{C}\) by radiation. The Newton's cooling coefficient is \(20 \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}\) and the Stefan-Boltzmann constant is \(5.67 \times 10^{-8}\) \(\mathrm{W} / \mathrm{m}^2 \mathrm{~K}^4\). If the rate of heat loss by convection is equal to the rate of heat loss by radiation, the emissivity of the body surface is
- A 0.35
- B 0.46
- C 0.55
- D 0.66
Answer & Solution
Correct Answer
(D) 0.66
Step-by-step Solution
Detailed explanation
Given, rate of heat lost by convection = Rate of heat lost by radiation…
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