Answer:
The heat flux between the surface of the pond and the surrounding air is 60 W/[tex]m^{2}[/tex]
Explanation:
Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;
q = Q/A ...............................1
since we are working with the convectional heat transfer coefficient equation 1 become;
q = h ([tex]T_{sf} -T_{sd}[/tex]) ........................2
where q is the heat flux;
Q is the heat energy that will be transferred;
h is the convectional heat coefficient = 20 W/[tex]m^{2}[/tex].K;
[tex]T_{sf}[/tex] is the surface temperature = [tex]23^{o}[/tex]C 23°C + 273.15 = 296.15 K;
[tex]T_{sd}[/tex] is the surrounding temperature = [tex]20^{o}[/tex]C = 20°C + 273.15 = 293.15 K;
The values are substituted into equation 2;
q = 20 W/[tex]m^{2}[/tex].K ( 296.15 K - 293.15 K)
q = 20 W/[tex]m^{2}[/tex].K ( 3 K)
q = 60 W/[tex]m^{2}[/tex]
Therefore the heat flux between the surface of the pond and the surrounding air is 60 W/[tex]m^{2}[/tex]
