Q52P - Fundamentals Of Physics

Question

Measurements in mines and boreholes indicate that Earth’s interior temperature increases with depth at the average rate of $\mathbf{30}\mathbf{°}\mathbf{ }\mathbf{C}\mathbf{/}\mathbf{km}$ . Assuming a surface temperature of $\mathbf{10}\mathbf{°}\mathbf{C}$ , at what depth does iron cease to be ferromagnetic? (The Curie temperature of iron varies very little with pressure.)

 

Step-by-Step Solution

Answer

The depth at which iron does cease to be ferromagnetic is  $\mathrm{d}=25\mathrm{km}$.

Step 1: Listing the given quantities

Temperature gradient is  $\frac{\mathrm{\Delta T}}{\mathrm{\Delta d}}=\frac{{30}^0\mathrm{C}}{\mathrm{km}}$ 

Surface temperature is  ${\mathrm{T}}_{\mathrm{s}}={10}^0\mathrm{C}$ 

Step 2: Understanding the concepts of temperature gradient

We use the concept of temperature gradient. Using the equation of temperature gradient and integrating it over the limit from the surface temperature to Curie temperature, we can find the depth at which iron ceases to be ferromagnetic.

Formulae:

 $\frac{\mathrm{\Delta T}}{\mathrm{\Delta d}}=\frac{{30}^0\mathrm{C}}{\mathrm{km}}$

Step 3: Calculations of the depth at which iron does cease to be ferromagnetic

We use the equation of temperature gradient.

 $\frac{\mathrm{\Delta T}}{\mathrm{\Delta d}}=\frac{{30}^0\mathrm{C}}{\mathrm{km}}$

We can rearrange it for $\mathrm{\Delta d}$ we get

 $\mathrm{\Delta d}=\frac{\mathrm{\Delta T}}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}$

We integrate it to find d. We get

 $\int \mathrm{\Delta d}=\underset{{\mathrm{T}}_{\mathrm{s}}}{\overset{{\mathrm{T}}_{\mathrm{c}}}{}}\frac{\mathrm{\Delta T}}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}$

Where,  ${\mathrm{T}}_{\mathrm{c}}$ - is Curie temperature.

$\begin{array}{c}\mathrm{d}=\frac{1}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}\underset{{\mathrm{T}}_{\mathrm{s}}}{\overset{{\mathrm{T}}_{\mathrm{c}}}{}}\mathrm{\Delta T}\\ =\frac{1}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}{\left|\mathrm{T}\right|}_{{\mathrm{T}}_{\mathrm{s}}}^{{\mathrm{T}}_{\mathrm{c}}} \\ =\frac{1}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}({\mathrm{T}}_{\mathrm{c}}-{\mathrm{T}}_{\mathrm{s}})\\ =\frac{1}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}({770}^0\mathrm{C}-{10}^0\mathrm{C})\\ =\frac{{760}^0\mathrm{C}}{\left(\frac{{30}^0\mathrm{C}}{\mathrm{km}}\right)}\\ =25.3\\ =25 \mathrm{km}\end{array}$

The depth at which iron does cease to be ferromagnetic is $\mathrm{d}=25\mathrm{km}$.