Sphere 1 with a diameter of 0.1 m is completely enclosed by another sphere 2 of diameter 0.4 m. The view factor F₁₂ is

The correct answer is 1.0.

To find the view factor $F_{12}$, we can analyze the geometry of the two concentric spheres:

1. Sphere 1 (Inner Sphere): Has a diameter of 0.1 m and is completely enclosed by Sphere 2.
Because Sphere 1 is a convex surface, it cannot "see" any part of itself. Consequently, any radiation leaving Sphere 1 cannot strike Sphere 1. Thus, the self-view factor is:
$F_{11}=0$

2. Summation Rule: For any surface $i$ inside an enclosure of $N$ surfaces, the sum of the view factors must equal 1:
$\sum _{j=1}^N{F}_{ij}=1$

Applying this summation rule to Sphere 1 in the two-surface enclosure (consisting of Sphere 1 and Sphere 2):
$F_{11}+F_{12}=1$

3. Calculation:
Substituting $F_{11}=0$ into the equation:
$0+F_{12}=1$
$F_{12}=1.0$

Since Sphere 1 is completely enclosed by Sphere 2, 100% of the radiation leaving Sphere 1 must intercept Sphere 2. Therefore, the view factor $F_{12}$ is exactly 1.0.