Solved Problem on Thermal Expansion

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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 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 (\( \alpha _{\;\text{v}} \))
Container with a volume of 125 cubic centimeters initially at a temperature of 20 degrees Celsius full of mercury and after heated to a temperature of 28 degrees Celsius and a Delta V ap apparent volume flowing out from the vessel.
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
\[ \bbox[#99CCFF,10px] {\Delta V_{\text{ap}}=V_{0}\gamma _{\text{ap}}\Delta t} \]
\[ \Delta V_{\text{ap}}=V_{0}\gamma_{\text{ap}}\;(\;t_{\text{f}}-t_{\text{i}}\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 10^{-6}.(\;28-20\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 8 \times 10^{-6} \]
\[ \bbox[#FFCCCC,10px] {\Delta V_{\text{ap}}=0.153\;\text{cm}^{3}} \]

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