Q.Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57$^\circ$C is drunk. You can take body (tooth) temperature to be 37$^\circ$C and $\alpha = 1.7 \times 10^{-5}$/$^\circ$C, bulk modulus for copper $= 140 \times 10^{9}$ N/m$^2$.

The key idea is thermal stress, which develops when a material is prevented from expanding or contracting due to a temperature change. For a material constrained in all directions, like copper in a tooth cavity, the stress is related to the volumetric thermal expansion and the Bulk Modulus.

1. First, calculate the change in temperature: $\Delta T = 57^\circ\text{C} - 37^\circ\text{C} = 20^\circ\text{C}$
2. The volumetric coefficient of thermal expansion ($\gamma$) is approximately three times the linear coefficient ($\alpha$): $\gamma = 3\alpha$.
3. The volumetric strain ($\frac{\Delta V}{V}$) due to this temperature change is $\gamma \Delta T = 3\alpha \Delta T$.
4. The stress ($P$) developed is then given by the product of the Bulk Modulus ($K$) and the volumetric strain: …

When a material is heated, its constituent particles vibrate more vigorously, causing the average distance between them to increase. This results in an increase in the material's dimensions, a phenomenon known as thermal expansion. If this expansion is allowed to occur freely, no internal stress develops. However, if the material is constrained and prevented from expanding, internal forces are generated within it. These forces, distributed over the material's cross-sectional area, constitute thermal stress.

In this problem, a copper filling in a tooth cavity is heated by hot tea. The copper attempts to expand, but the surrounding tooth cavity resists this expansion. This resistance leads to the development of compressive stress within the copper filling. Since the expansion is volumetric (the copper tries to occupy more space), the bulk modulus, which relates volumetric stress (pressure) to volumetric strain, is the appropriate elastic modulus to use.

1. Identify Given Values and Calculate Temperature Change: We are given the initial temperature (body/tooth temperature) $T_1 = 37^\circ \text{C}$ and the final temperature (hot tea temperature) $T_2 = 57^\circ \text{C}$. The change in temperature, $\Delta T$, is:

We are also given the coefficient of linear expansion for copper, $\alpha = 1.7 \times 10^{-5} /^\circ \text{C}$, and the bulk modulus for copper, $K = 140 \times 10^9 \text{ N/m}^2$.

2. Determine Volumetric Thermal Expansion:

If the copper were free to expand, its volume would increase due to the temperature rise. The fractional change in volume (volumetric strain) due to thermal expansion is given by:

where $V_0$ is the original volume and $\gamma$ is the coefficient of volumetric expansion. For an isotropic material like copper, the coefficient of volumetric expansion is approximately three times the coefficient of linear expansion:

Substituting the given value of $\alpha$:

So, the volumetric strain that *would have occurred* if the copper were free to expand is:

3. Relate Prevented Volumetric Strain to Stress using Bulk Modulus: The tooth cavity prevents this volumetric expansion. The bulk modulus $K$ is defined as the ratio of volumetric stress (pressure, $P$) to volumetric strain ($\Delta V/V_0$): …

Collapse the derivation into one formula: combining $\gamma=3\alpha$ with $K=\dfrac{P}{\Delta V/V_0}$ gives $P=3\alpha K\Delta T$ directly. Substituting: $P=3(1.7\times10^{-5})(140\times10^9)(20)=1.428\times10^8$ N/m², matching the step-by-step result in one line. Order-of-magnitude sanity check: $\alpha\Delta T\sim3\times10^{-4}$ is a tiny strain, but copper's bulk modulus is enormous ($\sim10^{11}$ Pa), so even this tiny prevented strain produces a stre …