Answer:
[tex]0.015\ \text{W}[/tex]
Explanation:
[tex]T_{set}[/tex] = Set temperature = [tex]70^{\circ}\text{C}[/tex]
[tex]T_\infty[/tex] = Air temperature = [tex]50^{\circ}\text{C}[/tex]
A = Surface area = [tex]30\ \text{mm}^2=30\times 10^{-6}\ \text{m}^2[/tex]
h = Convection heat transfer coefficient = [tex]25\ \text{W/m}^2\text{K}[/tex]
Heater power is given by
[tex]P_e=hA(T_{set}-T_\infty)\\\Rightarrow P_e=25\times 30\times 10^{-6}(70-50)\\\Rightarrow P_e=0.015\ \text{W}[/tex]
The required heater power is [tex]0.015\ \text{W}[/tex]
