Answer:
The surface temperature of the metal foil is 453.645 K
Step-by-step explanation:
The properties of the air at 100°C and 1 atm are the follows:
ρ = density = 0.9458 kg/m³
Cp = specific heat = 1009 J/kg K
Pr = 0.7111
The surface area of the plate is equal to:

Where
W = width of the plate = 0.2 m
L = length of the plate = 0.5 m

The average friction coefficient is equal to:

Where
F = drag force = 0.3 N
V = velocity of air flow = 100 m/s

The average heat transfer coefficient is equal to:

The surface temperature is:

Where
q = heat flux = 6100 W/m²
Tα = temperature of the air = 20°C = 293 K
