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The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J?
Most Upvoted Answer
The effective area of a black body is 0.1 m^2 and its temperature is 1...
Given:
Effective area of the black body = 0.1 m²
Temperature of the black body = 1000 K
To find:
Amount of radiations emitted by the black body per min
Solution:
According to Stefan's law, the amount of energy radiated per unit time (P) by a black body is given by:
P = σA(T^4)
Where,
σ = Stefan-Boltzmann constant = 5.67 × 10^-8 W/m²K^4
A = effective area of the black body
T = temperature of the black body
Substituting the given values, we get:
P = 5.67 × 10^-8 × 0.1 × (1000^4)
P = 5.67 × 10^-8 × 0.1 × 10^12
P = 56.7 × 10^3 W
Converting the power to energy per minute:
Energy emitted per minute = power × time
Energy emitted per minute = 56.7 × 10^3 × 60
Energy emitted per minute = 3.402 × 10^6 J/min
Converting to kcal/min:
1 kcal = 4.184 kJ
Therefore, 3.402 × 10^6 J/min = (3.402 × 10^6) / (4.184 × 10^3) kcal/min
Energy emitted per minute = 0.812 kcal/min
Answer:
The amount of radiations emitted by the black body per min is 0.812 kcal/min.
Hence, option (3) 5.63 kcal is incorrect. Option (2) 81 kcal is too high. Option (1) 1.34 kcal and option (4) 1.34 kJ are also incorrect.
Effective area of the black body = 0.1 m²
Temperature of the black body = 1000 K
To find:
Amount of radiations emitted by the black body per min
Solution:
According to Stefan's law, the amount of energy radiated per unit time (P) by a black body is given by:
P = σA(T^4)
Where,
σ = Stefan-Boltzmann constant = 5.67 × 10^-8 W/m²K^4
A = effective area of the black body
T = temperature of the black body
Substituting the given values, we get:
P = 5.67 × 10^-8 × 0.1 × (1000^4)
P = 5.67 × 10^-8 × 0.1 × 10^12
P = 56.7 × 10^3 W
Converting the power to energy per minute:
Energy emitted per minute = power × time
Energy emitted per minute = 56.7 × 10^3 × 60
Energy emitted per minute = 3.402 × 10^6 J/min
Converting to kcal/min:
1 kcal = 4.184 kJ
Therefore, 3.402 × 10^6 J/min = (3.402 × 10^6) / (4.184 × 10^3) kcal/min
Energy emitted per minute = 0.812 kcal/min
Answer:
The amount of radiations emitted by the black body per min is 0.812 kcal/min.
Hence, option (3) 5.63 kcal is incorrect. Option (2) 81 kcal is too high. Option (1) 1.34 kcal and option (4) 1.34 kJ are also incorrect.
Community Answer
The effective area of a black body is 0.1 m^2 and its temperature is 1...
Given that,
Temperature T = 1000 K
Surface area, A = 0.1 m²
Relative emittance, e = 1 (it varies from 0 to 1) Taken as 1.
Using Stefan's law of radiation,
The total heat energy emitted from a surface is proportional to the fourth power of temperature.
Formula for the energy radiated is, E = eσAT⁴
Where, is the Stefan's constant = 5.67 ×10⁻⁸
Then, substitute the values,
E = 1 × 5.67 × 10⁻⁸ × (1000)⁴ × 0.1
= 5670 W
1W = 0.0143 kcal/min
5670 W = 5670 × 0.0143 = 81Kcal/min
Temperature T = 1000 K
Surface area, A = 0.1 m²
Relative emittance, e = 1 (it varies from 0 to 1) Taken as 1.
Using Stefan's law of radiation,
The total heat energy emitted from a surface is proportional to the fourth power of temperature.
Formula for the energy radiated is, E = eσAT⁴
Where, is the Stefan's constant = 5.67 ×10⁻⁸
Then, substitute the values,
E = 1 × 5.67 × 10⁻⁸ × (1000)⁴ × 0.1
= 5670 W
1W = 0.0143 kcal/min
5670 W = 5670 × 0.0143 = 81Kcal/min
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The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J?
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The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J?.
The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The effective area of a black body is 0.1 m^2 and its temperature is 1000 K . The amount of radiations emitted by it per min is - (1) 1.34 kcal (2) 81kcal (3)5.63 kcal (4) 1.34 k J?.
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