A heat source at $$T = 10^3$$ K is connected to another heat reservoir at $$T = 10^2$$ K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 W K$$^{-1}$$ m$$^{-1}$$, the energy flux through it in the steady-state is:

We recall the steady-state heat conduction law, also called Fourier’s law of thermal conduction. It states that the energy flux $$\phi$$ (heat current density) through a homogeneous slab is given by

$$\phi \;=\; k \,\dfrac{\Delta T}{L}$$

where $$k$$ is the thermal conductivity of the material, $$\Delta T$$ is the temperature difference between the two faces of the slab, and $$L$$ is the thickness of the slab.

For the present problem we have the following data:

Thermal conductivity: $$k = 0.1\; \text{W K}^{-1}\text{ m}^{-1}$$

Temperatures of the reservoirs: $$T_\text{hot} = 10^{3}\; \text{K}$$ and $$T_\text{cold} = 10^{2}\; \text{K}$$

Therefore the temperature difference is

$$\Delta T = T_\text{hot} - T_\text{cold} = 10^{3} - 10^{2} = 1000 - 100 = 900\ \text{K}$$

The thickness of the copper slab is given as $$L = 1\ \text{m}$$.

Now we substitute these values into Fourier’s law:

$$\phi = k \,\dfrac{\Delta T}{L} = 0.1\;\text{W K}^{-1}\text{ m}^{-1} \times \dfrac{900\ \text{K}}{1\ \text{m}}$$

Carrying out the multiplication and division step by step, we first note that dividing by 1 m leaves the numerator unchanged in magnitude and only attaches the correct units:

$$\phi = 0.1 \times 900\ \text{W m}^{-2}$$

Multiplying $$0.1$$ and $$900$$ gives

$$\phi = 90\ \text{W m}^{-2}$$

Thus the steady-state energy flux through the copper slab is $$90\ \text{W m}^{-2}$$.

Looking at the provided options, this value corresponds to Option C.

Hence, the correct answer is Option C.