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A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K is
  • a)
    300 K
  • b)
    358 K
  • c)
    410 K
  • d)
    467 K
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A solar flux of intensity directly strikes a space vehicle surface whi...
For equilibrium,
Rate of heat absorbed = Rate of heat emitted
As the space vehicle is radiating to space at 0 K
∴ αAGr = ∊ A σ T14
⇒ 0.4 x 1400 = 0.6 x 5.67 x 10-8 x T14
⇒ T1 = 358.2 K
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Most Upvoted Answer
A solar flux of intensity directly strikes a space vehicle surface whi...
Calculation of Equilibrium Temperature

Given parameters:

- Absorptivity of the space vehicle surface = 0.4
- Emissivity of the space vehicle surface = 0.6

Step 1: Calculation of Solar Flux Intensity

As the solar flux of intensity directly strikes the space vehicle surface, we need to calculate its intensity first. Let's assume the intensity of solar flux as 'I'.

Step 2: Calculation of Absorbed Solar Flux

The space vehicle surface absorbs a fraction of the incident solar flux, which is equal to its absorptivity. Hence, the absorbed solar flux is given by:

Absorbed Solar Flux = Absorptivity x Solar Flux Intensity
= 0.4 x I

Step 3: Calculation of Radiative Heat Transfer

The space vehicle surface also radiates heat due to its emissivity. The radiative heat transfer from the surface is given by:

Radiative Heat Transfer = Emissivity x Stefan-Boltzmann Constant x (Equilibrium Temperature)^4
= 0.6 x 5.67 x 10^-8 x (Equilibrium Temperature)^4

Step 4: Calculation of Equilibrium Temperature

At equilibrium, the absorbed solar flux is equal to the radiative heat transfer. Hence, we can equate the expressions obtained in Step 2 and Step 3 and solve for the equilibrium temperature:

0.4 x I = 0.6 x 5.67 x 10^-8 x (Equilibrium Temperature)^4

Equilibrium Temperature = (0.4 x I / (0.6 x 5.67 x 10^-8))^0.25

Step 5: Calculation of Equilibrium Temperature for I = 1366 W/m2

The solar constant, which is the average solar flux intensity at the Earth's distance from the Sun, is approximately 1366 W/m2. Hence, we can substitute this value for 'I' in the expression obtained in Step 4 and solve for the equilibrium temperature:

Equilibrium Temperature = (0.4 x 1366 / (0.6 x 5.67 x 10^-8))^0.25
= 300 K

Therefore, the equilibrium temperature of the space vehicle surface in space at 0 K is 300 K (option A).
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A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K isa)300 Kb)358 Kc)410 Kd)467 KCorrect answer is option 'A'. Can you explain this answer?
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A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K isa)300 Kb)358 Kc)410 Kd)467 KCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K isa)300 Kb)358 Kc)410 Kd)467 KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solar flux of intensity directly strikes a space vehicle surface which has an absorptivity of 0.4 and emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K isa)300 Kb)358 Kc)410 Kd)467 KCorrect answer is option 'A'. Can you explain this answer?.
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