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Question: Coefficient of emission of an opaque body is \(0.6\). Radiant heat energy is incident on it at the r...

Coefficient of emission of an opaque body is 0.60.6. Radiant heat energy is incident on it at the rate of20watts20watts. Determine the quantity of radiant heat reflected by it in 22minutes​.
A. 480J480J
B. 960J960J
C. 1920J1920J
D. 1500J1500J

Explanation

Solution

In order to solve this question, we need to find the emissivity of the body given. Emissivity is the measure of degree to be able to radiate energy by a body. Use this concept to find the answer to this question.

Formula used:
E=εPtE = \varepsilon Ptwhere ε=\varepsilon = emissivity, P=P = Power, t=t = time taken

Complete answer:
Emissivity, ε=10.6\varepsilon = 1 - 0.6 \therefore because it is an opaque body.
ε=0.4\varepsilon = 0.4
We have provided P=P = 20watts20watts, t=2t = 2minutes and ε=0.4\varepsilon = 0.4
Substituting all these value in below formula we get energy:
E=εPt E=0.4×20×2×60 E=960J E = \varepsilon Pt \\\ E = 0.4 \times 20 \times 2 \times 60 \\\ E = 960J \\\ (convert minutes into seconds by multiplying it by 6060)

The radiant heat energy is option(B) 960J960J

Note: Points to be kept in mind while solving these questions:
Should have proper knowledge of emissivity because emissivity can be confusing sometimes.
Should learn all the important formulas as this topic contains lots of formulas which can be messy.