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Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^{\circ} \mathrm{C}$ is drunk. You can take body (tooth) temperature to be $37^{\circ} \mathrm{C}$ and $\alpha$ $=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$, bulk modulus for copper $=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$.
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As given that, change in temperature
$$
(\Delta t)=57-37=20^{\circ} \mathrm{C}
$$
Coefficient of linear expansion $(\alpha)$ of tooth body $=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Bulk modulus for copper $(B)=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Coefficient of cubical expansion $(\gamma)$ $=3 \alpha=3 \times 1.7 \times 10^{-5}=5.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Let the initial volume of the cavity be $V$ and its volume increases by $\Delta V$ due to increase in temperature $\Delta t$.
$$
\therefore \Delta V=\gamma V \Delta t
$$
Volumetric strain $\frac{\Delta V}{V}=\gamma \Delta t$
Thermal stress produced $=B \times$ Volumetric strain
$$
=B \times \frac{\Delta V}{V}=B \times \gamma \Delta t
$$
Thermal stress $=140 \times 10^9 \times\left(5.1 \times 10^{-5} \times 20\right)$
$$
=14280 \times 10^4=1.428 \times 10^8 \mathrm{~N} / \mathrm{m}^2
$$
This stress is about $10^3$ times of atmospheric pressure i.e., $1.01 \times 10^5 \mathrm{Nm}^{-2}$
$$
(\Delta t)=57-37=20^{\circ} \mathrm{C}
$$
Coefficient of linear expansion $(\alpha)$ of tooth body $=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Bulk modulus for copper $(B)=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Coefficient of cubical expansion $(\gamma)$ $=3 \alpha=3 \times 1.7 \times 10^{-5}=5.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Let the initial volume of the cavity be $V$ and its volume increases by $\Delta V$ due to increase in temperature $\Delta t$.
$$
\therefore \Delta V=\gamma V \Delta t
$$
Volumetric strain $\frac{\Delta V}{V}=\gamma \Delta t$
Thermal stress produced $=B \times$ Volumetric strain
$$
=B \times \frac{\Delta V}{V}=B \times \gamma \Delta t
$$
Thermal stress $=140 \times 10^9 \times\left(5.1 \times 10^{-5} \times 20\right)$
$$
=14280 \times 10^4=1.428 \times 10^8 \mathrm{~N} / \mathrm{m}^2
$$
This stress is about $10^3$ times of atmospheric pressure i.e., $1.01 \times 10^5 \mathrm{Nm}^{-2}$
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