Answer:
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Explanation:
As we know that heat flux is given by the formula
[tex]q^n = K\frac{dT}{dx}[/tex]
here we know that
K = thermal conductivity
[tex]\frac{dT}{dx}[/tex] = temperature gradient
now we know that
[tex]q^n = 11 W/m^2[/tex]
also we know that
K = 1.7 W/mK
now we have
[tex]11 = 1.7 \frac{dT}{dx}[/tex]
so temperature gradient is given as
[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]
also in other unit it will be same
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards