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A filament of 75 W light bulb may be considered as a black body radiating into a black enclosure at . The filament diameter is 0.1 mm and length is 5 cm. Considering only radiation, the filament temperature is
- a)3029 K
- b)2039 K
- c)2756 K
- d)1242 K
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A filament of 75 W light bulb may be considered as a black body radiat...
At steady state electric power supply to the bulb is equal to net radiation heat exchange between bulb and enclosure
here ∈1 = ∈2 = 1 and
F12 = 1; filament is completely enclosed
Q = A1F12σ(T14 - T24)
Q = (πDL) x 1 x 5.67 x 10-8 (T14 - 3434)
= 75 = 3.14 x 0.0001 x 0.05 x 1 x 5.67 x 10-8 (T14= 3434)
T1 = 3029.4 K
Most Upvoted Answer
A filament of 75 W light bulb may be considered as a black body radiat...
To determine the filament temperature of a 75 W light bulb, we can consider it as a black body radiating into a black enclosure.
Let's break down the solution into steps:
Step 1: Determine the surface area of the filament
The surface area of the filament can be calculated using the formula for the surface area of a cylinder:
Surface Area = 2πrL + πr^2
Given that the diameter of the filament is 0.1 mm, the radius (r) can be calculated as 0.05 mm or 0.00005 m. The length (L) of the filament is 5 cm or 0.05 m.
Substituting the values into the formula, we get:
Surface Area = 2π(0.00005)(0.05) + π(0.00005)^2
Surface Area ≈ 0.0000157 m^2
Step 2: Calculate the power radiated by the filament
The power radiated by a black body can be calculated using the Stefan-Boltzmann law:
Power = σ * A * T^4
Where σ is the Stefan-Boltzmann constant (approximately equal to 5.67 x 10^-8 W/m^2K^4), A is the surface area of the filament, and T is the temperature of the filament.
Given that the power of the light bulb is 75 W and the surface area of the filament is 0.0000157 m^2, we can rearrange the formula to solve for T:
T^4 = Power / (σ * A)
T^4 = 75 / (5.67 x 10^-8 * 0.0000157)
T^4 ≈ 3029
Taking the fourth root of both sides, we get:
T ≈ 3029 K
Therefore, the temperature of the filament is approximately 3029 K.
Let's break down the solution into steps:
Step 1: Determine the surface area of the filament
The surface area of the filament can be calculated using the formula for the surface area of a cylinder:
Surface Area = 2πrL + πr^2
Given that the diameter of the filament is 0.1 mm, the radius (r) can be calculated as 0.05 mm or 0.00005 m. The length (L) of the filament is 5 cm or 0.05 m.
Substituting the values into the formula, we get:
Surface Area = 2π(0.00005)(0.05) + π(0.00005)^2
Surface Area ≈ 0.0000157 m^2
Step 2: Calculate the power radiated by the filament
The power radiated by a black body can be calculated using the Stefan-Boltzmann law:
Power = σ * A * T^4
Where σ is the Stefan-Boltzmann constant (approximately equal to 5.67 x 10^-8 W/m^2K^4), A is the surface area of the filament, and T is the temperature of the filament.
Given that the power of the light bulb is 75 W and the surface area of the filament is 0.0000157 m^2, we can rearrange the formula to solve for T:
T^4 = Power / (σ * A)
T^4 = 75 / (5.67 x 10^-8 * 0.0000157)
T^4 ≈ 3029
Taking the fourth root of both sides, we get:
T ≈ 3029 K
Therefore, the temperature of the filament is approximately 3029 K.
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A filament of 75 W light bulb may be considered as a black body radiating into a black enclosure at . The filament diameter is 0.1 mm and length is 5 cm. Considering only radiation, the filament temperature isa)3029 Kb)2039 Kc)2756 Kd)1242 KCorrect answer is option 'A'. Can you explain this answer?
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