Solved Problem on Thermal Expansion Português English

A container is filled with 125 cm3 of mercury at a temperature of 20 °C. The average coefficient of expansion of the mercury is 180 × 10−6 °C−1 and the coefficient of linear expansion of the glass is 9 × 10−6 °C−1. Find the volume of mercury spilled out when the temperature rises to 28 °C.

Problem data
• volume of the container at 20°C:    V0 = 125 cm3;
• coefficient of expansion of mercury:    γHg = 180 × 10−6 °C−1;
• coefficient of linear expansion of glass:    αv = 9 × 10−6 °C−1;
• initial system temperature:    ti = 20 °C;
• final system temperature:    tf = 28 °C.
Problem diagram

At the initial temperature, the container 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 Δ Vap we want to find. figure 1

Solution

The coefficient of volumetric expansion of glass (γv) obtained from the coefficient of linear expansion of glass (αv)
$\bbox[#99CCFF,10px] {\gamma _{\text{v}}=3\alpha _{\text{v}}}$
$\gamma _{\text{v}}=3 \times 9 \times 10^{-6}\\ \gamma_{\text{v}}=27 \times 10^{-6\;\text{o}}\text{C}^{-1}$
The apparent coefficient of expansion of mercury, it is 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
$\bbox[#99CCFF,10px] {\Delta V_{\text{ap}}=V_{0} \gamma _{\text{ap}} \Delta t}$
$\Delta V_{\text{ap}}=V_{0}\gamma_{\text{ap}}\;(\;t_{\text{f}}-t_{\text{i}}\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 10^{-6} \times (\;28-20\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 8 \times 10^{-6}$
$\bbox[#FFCCCC,10px] {\Delta V_{\text{ap}}=0.153\;\text{cm}^{3}}$ 