To solve this problem we will use the two principles that are visible according to the phenomena described in the problem: Heat transfer by conductivity and Heat transfer by convection.
This thermal transfer will be equivalent and with it we can find the value asked.
Note: We will assume that the temperature value at the plate surface is: 60 ° C (For the given value of 650 50)
For Thermal Transfer by Conduction
[tex]Q_{cn} = -kA \frac{\Delta T}{\Delta x}[/tex]
[tex]Q_{cn} = -kA \frac{T_1-T_2}{L}[/tex]
Where,
k = Thermal conductivity
A = Cross-sectional Area
[tex]T_2[/tex] = Temperature of the bottom surface
[tex]T_1[/tex]= Temperature of the top surface
L = Length
Replacing we have that
[tex]Q_{cn} = -(80W\cdot K)(A)\frac{50\°C-60\°C}{15cm\frac{1m}{100cm}}[/tex]
[tex]Q_{cn} = 5333.33A[/tex]
For Thermal Transfer by Convection
[tex]Q_{cv} = hA(T_1-T_{\infty})[/tex]
Where,
h = Convection heat transfer coefficient
[tex]T_{\infty}[/tex]= Surrounding temperature
A = Surface Area
Replacing we have that
[tex]Q_{cv} = hA(50\°C-10\°C)[/tex]
[tex]Q_{cv} = 40hA[/tex]
Since the rate of heat transfer by convection is equal to that given by conduction we have to:
[tex]Q_{cn}=Q_{cv}[/tex]
[tex]5333.33A = 40hA[/tex]
[tex]h = 133.33W/m^2\cdot K[/tex]
It is stated that the typical values of forced convection of gases lies in the range of [tex](25-250)W/m^2\cdot K[/tex]. The obtained value is reasonable for forced convection of air.
