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A plate having 10 cm2 area each side is hanging in the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature and emissivity values for the surface of the room are 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heat loss from the two surface of the plate is
- a)13.66 W
- b)27.32 W
- c)27.87 W
- d)13.66 MW
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A plate having 10 cm2 area each side is hangingin the middle of a room...
A1 = 20 cm2 = 20 x 10-4m2
A2 = 100 m2
T1 = 800K
∈1 = 0.6
T2 = 300 K
∈2 = 0.3
A2 = 100 m2
T1 = 800K
∈1 = 0.6
T2 = 300 K
∈2 = 0.3
Most Upvoted Answer
A plate having 10 cm2 area each side is hangingin the middle of a room...
The Stefan-Boltzmann law relates the radiated power (P) from an object to its temperature (T) and emissivity (ε) as:
P = εσA(T^4)
Where:
P = radiated power
ε = emissivity
σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4))
A = surface area
T = temperature
Given:
Plate area (A_plate) = 10 cm^2 = 0.001 m^2
Room area (A_room) = 100 m^2
Plate temperature (T_plate) = 800 K
Plate emissivity (ε_plate) = 0.6
Room temperature (T_room) = 300 K
Room emissivity (ε_room) = 0.3
To calculate the radiated power from the plate, we can use the Stefan-Boltzmann law:
P_plate = ε_plateσA_plate(T_plate^4)
P_plate = (0.6)(5.67 x 10^-8 W/(m^2K^4))(0.001 m^2)((800 K)^4)
P_plate ≈ 0.00245 W
To calculate the radiated power from the room, we can use the Stefan-Boltzmann law with the room temperature and emissivity:
P_room = ε_roomσA_room(T_room^4)
P_room = (0.3)(5.67 x 10^-8 W/(m^2K^4))(100 m^2)((300 K)^4)
P_room ≈ 135.36 W
Therefore, the radiated power from the plate is approximately 0.00245 W and the radiated power from the room is approximately 135.36 W.
P = εσA(T^4)
Where:
P = radiated power
ε = emissivity
σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4))
A = surface area
T = temperature
Given:
Plate area (A_plate) = 10 cm^2 = 0.001 m^2
Room area (A_room) = 100 m^2
Plate temperature (T_plate) = 800 K
Plate emissivity (ε_plate) = 0.6
Room temperature (T_room) = 300 K
Room emissivity (ε_room) = 0.3
To calculate the radiated power from the plate, we can use the Stefan-Boltzmann law:
P_plate = ε_plateσA_plate(T_plate^4)
P_plate = (0.6)(5.67 x 10^-8 W/(m^2K^4))(0.001 m^2)((800 K)^4)
P_plate ≈ 0.00245 W
To calculate the radiated power from the room, we can use the Stefan-Boltzmann law with the room temperature and emissivity:
P_room = ε_roomσA_room(T_room^4)
P_room = (0.3)(5.67 x 10^-8 W/(m^2K^4))(100 m^2)((300 K)^4)
P_room ≈ 135.36 W
Therefore, the radiated power from the plate is approximately 0.00245 W and the radiated power from the room is approximately 135.36 W.
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A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer?
Question Description
A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer?.
A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate having 10 cm2 area each side is hangingin the middle of a room of 100 m2 total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature andemissivity values for the surface of the roomare 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10-8 W/m2K4. The total heatloss from the two surface of the plate isa)13.66 W b)27.32 Wc)27.87 W d)13.66 MWCorrect answer is option 'B'. Can you explain this answer?.
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