Answer:
![T_(surface)=343.86\°c](https://img.qammunity.org/2020/formulas/engineering/college/5g6a5jauj089zyke5q1o46xy6eeprce77p.png)
Step-by-step explanation:
We define the constant, so
![k=0.0258W/m^2K\\\upsilon_(air)=1.608*10^5m^2/s\\Pr=0.7282](https://img.qammunity.org/2020/formulas/engineering/college/kir13137atdvhgnskb5zwy3adclw067myi.png)
So, begin calculating the Reynolds number, so
![(Re)_(x=1.5) = (V_(\infty)x)/(\upsilon_(air))\\(Re)_(x=1.5) = (2.5*1.5)/(1.608*10^(-5))}\\(Re)_(x=1.5) = 2.33*10^(5)](https://img.qammunity.org/2020/formulas/engineering/college/my50nk89sg59t2ktazm4g12wyxt89yqqxx.png)
How reynolds number is greater than critical reynolds number, the flow
of the air is near about the turbulent flow,
we now calculate the local nusselt number
![(Nu)_x = 0.332 (Re)^(1/2)_x(Pr)^(1/3)\\(Nu)_x= 0.322(2.33*10^5)^(1/2)*(0.782)^(1/3)\\(Nu)_x = 143.1977](https://img.qammunity.org/2020/formulas/engineering/college/2v9qjq3aekxc4xdooq0sdl4sq7e57rmyyx.png)
calculate the local convection heat transfer coefficient
![h_X = ((Nu)_xk)/(x)\\h_x= (143.1977*0.0258)/(1.5)\\h_x=2.46300 W/m^2K](https://img.qammunity.org/2020/formulas/engineering/college/ba56a2ar004mz8a9y6pcdusalfbecmmfq5.png)
Apply the energy balance equation
![q=h_x(T_(surface)-T_(\infty))\\810=2.463*(T_(surface)-15)](https://img.qammunity.org/2020/formulas/engineering/college/7wpaugz70xa38rdsmpfis6rt3hezaawcd7.png)
![T_(surface)=343.86\°c](https://img.qammunity.org/2020/formulas/engineering/college/5g6a5jauj089zyke5q1o46xy6eeprce77p.png)