Onsider a 20-cm-thick granite wall with a thermal conductivity of 2.79 w/m·k. the temperature of the left surface is held constant at 50°c, whereas the right face is exposed to a flow of 22°c air with a convection heat transfer coefficient of 15 w/m2·k. neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

q=heat flux=216.83W/m^2

Ts=36.45C

Explanation:

What you should do is raise the heat transfer equation from the inside of the granite wall to the air.

You should keep in mind that in the numerator you place the temperature differences, while the denominator places the sum of the thermal resistances by conduction and convection.

When you have the heat value, all you have to do is use the convection equation q = h (ts-t) and solve for the surface temperature.

I attached procedure

we need to know the time interval during which the force acts. impulse of force = momentum change