### Solved Problem on Thermal Expansion Português English
A vessel is completely filled with 125 cm3 of mercury at a temperature of 20°C. The average coefficient of expansion of the mercury is $$180 \times 10^{-6\;\text{o}}\text{C}^{-1}$$ and the coefficient of linear expansion of the glass is $$9 \times 10^{-6\;\text{o}}\text{C}^{-1}$$. Determine the volume of mercury spilled out when the temperature rises to 28°C.

Problem data
• volume of the vessel at 20°C:    $$V_{0}=125\;\text{cm}^{3}$$;
• coefficient of expansion of mercury:    $$\gamma _{\text{Hg}}=180 \times 10^{-6\;\text{o}}\text{C}^{-1}$$;
• coefficient of linear expansion of glass:    $$\alpha _{\text{v}}=9 \times 10^{-6\;\text{o}}\text{C}^{-1}$$;
• initial system temperature:    $$t_{\text{i}}=20\;^{\text{o}}\text{C}$$;
• final system temperature:    $$t_{\text{f}}=28\;^{\text{o}}\text{C}$$.
Problem sketch

At the initial temperature the vessel and the mercury have the same volume as 125 cm3, when the system is heated it expands, but as the mercury expands more than the container a part of it overflows. This overflowing part is the apparent volume change ($$\Delta V_{\text{ap}}$$) we wish to find.

Solution

The coefficient of volumetric expansion of the glass ($$\gamma _{\text{v}}$$) obtained from the coefficient of linear expansion of the glass ($$\alpha _{\;\text{v}}$$) figure 1
$\bbox[#99CCFF,10px] {\gamma _{\text{v}}=3\alpha _{\text{v}}}$
$\gamma _{\text{v}}=3 \times 9 \times 10^{-6}\\ \gamma_{\text{v}}=27 \times 10^{-6\;\text{o}}\text{C}^{-1}$
The coefficient of apparent expansion of the mercury, only of the part that overflowed, is given by
$\bbox[#99CCFF,10px] {\gamma _{\text{ap}}=\gamma _{\text{Hg}}-\gamma _{\text{v}}}$
$\gamma _{\text{ap}}=180 \times 10^{-6}-27 \times 10^{-6}\\ \gamma_{\text{ap}}=153 \times 10^{-6\acute{}\text{o}}\text{C}^{-1}$
The spilled volume will be
$\bbox[#99CCFF,10px] {\Delta V_{\text{ap}}=V_{0}\gamma _{\text{ap}}\Delta t}$
$\Delta V_{\text{ap}}=V_{0}\gamma_{\text{ap}}\;(\;t_{\text{f}}-t_{\text{i}}\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 10^{-6}.(\;28-20\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 8 \times 10^{-6}$
$\bbox[#FFCCCC,10px] {\Delta V_{\text{ap}}=0.153\;\text{cm}^{3}}$