Answer:
125 W
Explanation:
Data:
Let the Biot number be:
[tex]L_{c} = \frac{V}{A} = \\\frac{\pi /6D^{3} }{\pi D^{2} }\\ = \frac{D}{6} \\= \frac{1.2cm}{6} \\= 0.002m[/tex]
Then:
[tex]Bi = \frac{hL_{c} }{k} = 0.0166[/tex]
We know that:
L[tex]L_{c} = 0.002[/tex]
K= 15.1 W/m²K
Then, substituting gives:
125 W