Question
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 $\alpha=1.7 \times 10^{-5} / K$. Bulk modulus for copper $=140$ $\times 10^9 N / m ^2$.

Answer

First of all, the stress on the object(tooth) developed over here is dependent on bulk modulus of the material and the temperature difference. That's the stress is called a thermal stress over here.
Now, change in temperature $\Delta T=57-37=20^{\circ} C =20 K$ (since temperature difference is same here in both the scales)
Coefficient of linear expansion $\alpha$ of (tooth) body $=1.7 \times 10^{-5} K^{-1}$
Cubical expansion $\gamma=3 \alpha=3 \times 1.7 \times 10^{-5}=5.1 \times 10^{-5} K^{-1},\left[ As , \alpha=\frac{\gamma}{3}\right]$
Let the volume of the cavity be V and its volume increased by $\Delta V$ due to increase in temperature $\Delta T$.
Then from the defination of coefficient of volume expansion,
$
\begin{aligned}
& \Delta V=\gamma V \cdot \Delta T \\
& \Rightarrow \frac{\Delta V}{V}=\gamma \Delta T
\end{aligned}
$
Thermal stress produced $=$ Bulk modulus $\times$ Volume strain
$
=B \times \frac{\Delta V}{V}=B \times \gamma \times \Delta T
$
Hence, thermal stress $=140 \times 10^9 \times 5.1 \times 10^{-5} \times 20=14280 \times 10^4 N / m ^2$, as bulk modulus $=140 \times 10^9 N / m ^2($ given $)$
$
=1.428 \times 10^8 Nm^{-2}
$
This stress is about $10^3$ times of atmospheric pressure i.e., $1.013 \times 10^5 Nm ^{-2}$.

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