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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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmann's constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is
[2003]
- 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 hanging in the middle of a roo...
Heal loss from one surface of the plate,
Putting
A1 = 10 × 10–4 m2, T1 = 800
∈1 = 0.6 T2 = 300
A2 = 100 m2
∈2 = 0.3
= 13.66 W
∴ Heat loss from both the surfaces
= 2 Qnet
= 2 × 13.66 = 27.32 W
Putting
A1 = 10 × 10–4 m2, T1 = 800
∈1 = 0.6 T2 = 300
A2 = 100 m2
∈2 = 0.3
= 13.66 W∴ Heat loss from both the surfaces
= 2 Qnet
= 2 × 13.66 = 27.32 W
Most Upvoted Answer
A plate having 10 cm2 area each side is hanging in the middle of a roo...
Is 1.38 x 10^-23 J/K and the Stefan-Boltzmann constant is 5.67 x 10^-8 W/m^2K^4.
To calculate the heat transfer between the plate and the room, we can use the formula for radiative heat transfer:
Q = σεA(T1^4 - T2^4)
where Q is the heat transfer rate, σ is the Stefan-Boltzmann constant, ε is the emissivity, A is the surface area, T1 is the temperature of the plate, and T2 is the temperature of the room.
Substituting the given values, we get:
Q = (5.67 x 10^-8)(0.6)(10^-4)(800^4 - 300^4) = 291.7 W
Therefore, the heat transfer rate between the plate and the room is 291.7 W.
To calculate the heat transfer between the plate and the room, we can use the formula for radiative heat transfer:
Q = σεA(T1^4 - T2^4)
where Q is the heat transfer rate, σ is the Stefan-Boltzmann constant, ε is the emissivity, A is the surface area, T1 is the temperature of the plate, and T2 is the temperature of the room.
Substituting the given values, we get:
Q = (5.67 x 10^-8)(0.6)(10^-4)(800^4 - 300^4) = 291.7 W
Therefore, the heat transfer rate between the plate and the room is 291.7 W.
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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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)13.66 MWCorrect answer is option 'B'. Can you explain this answer?
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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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)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 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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)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 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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)13.66 MWCorrect answer is option 'B'. Can you explain this answer?.
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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)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 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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)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 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 valute for the surfaces of the room are 300 K and 0.3 respectively. Boltzmanns constant σ = 5.67 × 10-8 W/m2K4. The total heat loss from the two surfaces of the plate is[2003]a)13.66 Wb)27.32 Wc)27.87 Wd)13.66 MWCorrect answer is option 'B'. Can you explain this answer?.
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