Heat transfer from a hot cube to a constant-temperature bath: If the cube loses heat from the top, bottom, and the four side faces with surface heat transfer coefficients h1, h2, and h3 respectively, what is the area-averaged overall heat transfer coefficient for the entire cube?

Introduction / Context:
This question tests understanding of convective heat transfer from composite surfaces. A cube immersed in a constant-temperature bath can have different local heat transfer coefficients on different faces. The correct “average” coefficient must be area-weighted, not an arithmetic or geometric mean chosen blindly.

Given Data / Assumptions:

• Cubic body; each face has equal area A.
• Top face coefficient = h1; bottom face coefficient = h2; each of four side faces coefficient = h3.
• Surroundings are isothermal; film coefficients are uniform over each face.

Concept / Approach:
The overall or average heat transfer coefficient h_avg for multiple parallel heat transfer paths over distinct areas is the area-weighted mean. Total heat rate is the sum of facewise convection rates. Divide total by total area times the same driving temperature difference to define h_avg.

Step-by-Step Solution:

Verification / Alternative check:
If all faces had the same coefficient h, formula gives h_avg = (h + h + 4h)/6 = h, confirming consistency.

Why Other Options Are Wrong:

• h1 + h2 + h3 ignores multiplicity of side faces and total area.
• (h1h2h3)^(1/3) is an unjustified geometric mean; convection adds linearly over areas.
• 2(h1 + h2 + h3)/3 is an arbitrary average and miscounts side faces.

Common Pitfalls:
Forgetting that there are four side faces; using simple averages instead of area-weighted sums.

Final Answer:
(h1 + h2 + 4*h3) / 6