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A black body radiates heat energy at the rate of 2 x 105J/s-m2 at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2 will be
- a)200 K
- b)400 K
- c)600 K
- d)800 K
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K...
Given: Rate of heat radiation at temperature T1 = 2 x 10^5 J/s-m^2
Rate of heat radiation at temperature T2 = 32 x 10^5 J/s-m^2
We need to find the temperature T2.
We can use the Stefan-Boltzmann law which states that the rate of heat radiation from a black body is proportional to the fourth power of its absolute temperature.
Mathematically,
P = σAεT^4
Where P is the rate of heat radiation, A is the surface area of the black body, ε is the emissivity of the black body (which is 1 for a black body), σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2-K^4), and T is the absolute temperature.
Using this formula, we can write:
T2^4/T1^4 = (32 x 10^5)/(2 x 10^5)
Simplifying this equation, we get:
T2^4 = 256T1^4
Taking the fourth root on both sides, we get:
T2 = (256T1^4)^(1/4)
T2 = T1 x 4
Substituting the given values, we get:
T2 = 400 K x 4
T2 = 1600 K
Therefore, the temperature at which the rate of heat radiation is 32 x 10^5 J/s-m^2 is 1600 K, which is option D.
Rate of heat radiation at temperature T2 = 32 x 10^5 J/s-m^2
We need to find the temperature T2.
We can use the Stefan-Boltzmann law which states that the rate of heat radiation from a black body is proportional to the fourth power of its absolute temperature.
Mathematically,
P = σAεT^4
Where P is the rate of heat radiation, A is the surface area of the black body, ε is the emissivity of the black body (which is 1 for a black body), σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2-K^4), and T is the absolute temperature.
Using this formula, we can write:
T2^4/T1^4 = (32 x 10^5)/(2 x 10^5)
Simplifying this equation, we get:
T2^4 = 256T1^4
Taking the fourth root on both sides, we get:
T2 = (256T1^4)^(1/4)
T2 = T1 x 4
Substituting the given values, we get:
T2 = 400 K x 4
T2 = 1600 K
Therefore, the temperature at which the rate of heat radiation is 32 x 10^5 J/s-m^2 is 1600 K, which is option D.
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Community Answer
A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K...
E proportional to T^4
2×10^5/32×10^5 =400^4/t^4
ie, t=2×400=800
option d is answer
2×10^5/32×10^5 =400^4/t^4
ie, t=2×400=800
option d is answer
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A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer?.
A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A black body radiates heat energy at the rate of 2 x 105J/s-m2at 400 K. The temperature of the black body at which rate of heat radiation is 32 x 105 J/s-m2will bea)200 Kb)400 Kc)600 Kd)800 KCorrect answer is option 'D'. Can you explain this answer?.
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