Heat Transfer Questions and Answers – Gray Body and Selective Emitters

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This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Gray Body And Selective Emitters”.

1. Consider two bodies, one absolutely back and the other non-black and let these be at same temperature. Which one of the following statement is correct?
a) Black body radiates less intensively than a non-black body
b) Non-black body radiates less intensively than a black body
c) Both will radiates equally
d) None of them will radiates
View Answer

Answer: b
Explanation: The radiation spectrum for a non-black body may be similar or radically different from that of a black body.

2. When Stefan-Boltzmann law is applied to a black body, it takes the form
a) E = σ T
b) E = σ T 2
c) E = σ T 3
d) E = σ T 4
View Answer

Answer: d
Explanation: The constant is different for different bodies.

3. When the emissivity of non-black surface is constant at all temperatures and throughout the entire range of wavelength, the surface is called
a) Gray body
b) Transparent body
c) Opaque bodies
d) Perfect black body
View Answer

Answer: a
Explanation: The radiation spectrum for a grey body, though reduced in vertical scale, is continuous and identical to the corresponding curve for a perfectly black surface.
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4. The emissivity of the gray surface may be expressed as
a) σ /2 σ b
b) σ / 3 σ b
c) σ / σ b
d) ½ σ / σ b
View Answer

Answer: c
Explanation: It is the ratio of radiating coefficient to that of black body radiating coefficient.

5. If a black body at 1000 K and a gray body at 1250 K emit the same amount of radiation, what should be the emissivity of the gray body?
a) 0.3096
b) 0.4096
c) 0.5096
d) 0.6096
View Answer

Answer: b
Explanation: E = (T b/T G) 4 = 0.4096.
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6. The radiant heat transfer from a plate of 2.5 cm2 area at 1250 K to a very cold enclosure is 5.0 W. Determine the emissivity of the plate at this temperature
a) 0.444
b) 0.344
c) 0.244
d) 0.144
View Answer

Answer: d
Explanation: Emissivity = E/σ A T4 = 0.144.

7. A 100 W light bulb has a tungsten filament (emissivity = 0.30) which is required to operate at 2780 K. If the bulb is completely evacuated, calculate the minimum surface area of the tungsten filament
a) 0.98 * 10 -4 m2
b) 1.98 * 10 -4 m2
c) 2.98 * 10 -4 m2
d) 3.98 * 10 -4 m2
View Answer

Answer: a
Explanation: E = (Emissivity) σ A T4. So, A = 0.98 * 10 -4 m2.
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8. The monochromatic emissivity € of a diffuse surface at 1600 K varies with wavelength in the following manner
€ = 0.4 for 0 < λ < 2
= 0.8 for 2 < λ < 5
Determine the total emissivity
Find the total emissivity if monochromatic emissivity € diffuse surface at 1600 K
a) 0.5558
b) 0.5568
c) 0.5578
d) 0.5588
View Answer

Answer: c
Explanation: Total emissivity = 0.4 (-0.3181 – 0.0000) + 0.8 (0.8563 – 0.3118) = 0.5578.

9. For a hemisphere solid angle is measured as
a) Radian and its maximum value is π
b) Degree and its maximum value is 180 degree
c) Steradian and its maximum value is 2π
d) Steradian and its maximum value is π
View Answer

Answer: d
Explanation: It should be measured in Steradian.
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10. A gray body (E = 0.8) emits the same amount of heat as a black body at 1075 K. Find out the required temperature of the gray body
a) 1146.72 K
b) 1136.72 K
c) 1126.72 K
d) 1116.72 K
View Answer

Answer: b
Explanation: T b4 = E T g4.

Sanfoundry Global Education & Learning Series – Heat Transfer.

To practice all areas of Heat Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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