Heat Transfer Questions and Answers – Stefan- Boltzman Law

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This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Stefan- Boltzman Law”.

1. The Stefan-Boltzmann constant has units of
a) kcal/m2 hr K4
b) kcal/m hr K4
c) kcal/hr K4
d) kcal/m2 K4
View Answer

Answer: a
Explanation: According to Stefan-Boltzmann law, q = α A T4.

2. According to Stefan-Boltzmann law of thermal radiation
a) q = α A T
b) q = α A T4
c) q = α A T3
d) q = α A T5
View Answer

Answer: b
Explanation: α is Stefan-Boltzmann constant whose value is 5.67 * 10 -8 W/m2 K4.

3. Calculate the radiant flux density from a black surface at 400 degree Celsius?
a) 1631.7 W/m2
b) 31.7 W/m2
c) 631.7 W/m2
d) 11631.7 W/m2
View Answer

Answer: d
Explanation: E = σ T4 = 5.67 * 10 –8 (400 + 273)4 = 11631.7 W/m2.

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4. If the emitted radiant energy is to be doubled, to what temperature surface of black body needs to be raised? Take radiant flux density as 11631.7 W/m 2.
a) 894.4 K
b) 200.4 K
c) 800.3 K
d) 600.4 K
View Answer

Answer: d
Explanation: 2(11631.7) = 5.67 * 10 –8 T 4.

5. A furnace having inside temperature of 2250 has a glass circular viewing of 6 cm diameter. If the transmissivity of glass is 0.08, make calculations for the heat loss from the glass window due to radiation
a) 234.54 W
b) 652.32 W
c) 328.53 W
d) 762.32 W
View Answer

Answer: c
Explanation: Q = σ A T4 (t) = 328.53 W.

6. The value of radiation coefficient or the Stefan-Boltzmann constant is
a) 5.67 * 10 -8 W/m2 K4
b) 5.67 * 10 -7 W/m2 K4
c) 5.67 * 10 -6 W/m2 K4
d) 5.67 * 10 -5 W/m2 K4
View Answer

Answer: a
Explanation: q = α A T4.

7. Measurements were made of the monochromatic absorptivity and monochromatic hemispherical irradiation incident on an opaque surface, and the variation of these parameters with wavelength may be approximated by the result shown below. Determine the total hemispherical absorptivity
Find the total hemispherical absorptivity for variation of these parameters with wavelength
a) 0.557
b) 0.667
c) 0.777
d) 0.887
View Answer

Answer: b
Explanation: Incident flux = 800(8 – 2) – 4800 W/m2. Absorptivity = 3200/4800 = 0.667.
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8. What is the wavelength band for Ultraviolet rays?
a) 1 * 10 -6 to 3.9 * 10 -1 micron meter
b) 1 * 10 -4 to 3.9 * 10 -1 micron meter
c) 2 * 10 -3 to 3.9 * 10 -1 micron meter
d) 1 * 10 -2 to 3.9 * 10 -1 micron meter
View Answer

Answer: d
Explanation: This is the maximum and minimum wavelength for Ultraviolet rays.

9. A black body of total area 0.045 m2 is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m 2 and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black body
a) 547.3 K
b) 287.4 K
c) 955.9 K
d) 222.2 K
View Answer

Answer: c
Explanation: Q r = σ A (T b4 – Tw4), Q c = k A d t/δ = 1979.5 W. So temperature of black body is 955.9 K.
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10. What is the wavelength band for solar radiation?
a) 1 * 10 -1 to 3 micron meter
b) 1 * 10 -1 to 2 micron meter
c) 1 * 10 -1 to 1 micron meter
d) 1 * 10 -1 to 10 micron meter
View Answer

Answer: a
Explanation: This is the maximum and minimum wavelength for solar radiation.

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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