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GATE 2024 Chemical Engineering, Q37 · Heat Transfer

The correct answer is A, 759.

Question

The temperatures of two large parallel plates of equal emissivity are 900 K and 300 K. A reflective radiation shield of low emissivity and negligible conductive resistance is placed parallel between them. The steady-state temperature of the shield, in K, is

  • A. 759
  • B. 559
  • C. 659
  • D. 859

Official answer: 759

Why

GATE 2024 official key: (A). For a shield with the same emissivity on both faces, symmetry of the radiative network gives Tshield4=(T14+T24)/2=(9004+3004)/2,T_{shield}⁴ = (T1⁴+T2⁴)/2 = (900⁴+300⁴)/2, independent of the shield's emissivity value. This gives Tshield≈759T_{shield} \approx 759 K.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.

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