A container with an internal volume of 1 liter at 20 °C is heated to 100 °C. Find the volume of this container
after heating, knowing that the coefficient of linear expansion of the material is
15 × 10
−6 °C
−1.
Problem data:
- Initial internal volume: V0 = 1 ℓ;
- Initial temperature: ti = 20 °C;
- Final temperature: tf = 100 °C;
- Coefficient of linear expansion of the container: α = 15 × 10−6 °C−1.
Problem diagram:
The problem is equivalent to a body constructed of the same material as the container and of the same internal volume
V0 as the inner part of the container (Figure 1-A). When the body is heated, it will expand (Figure 1-B).
The heated body will reach a new volume
V after expansion (Figure 2-A)
The internal volume of the heated container will be the same volume
V as the body (Figure 2-B).
Solution
First, we convert the volume given in liters to cubic meters used in the
International System of Units (
S.I.)
\[
V=1\;\cancel{\ell} \times \frac{1\;\text{m}^{3}}{1000\;\cancel{\ell}}=\frac{1}{1000}\;\text{m}^{3}=\frac{1}{10^{3}}\;\text{m}^{3}=1 \times 10^{-3}\;\text{m}^{3}
\]
The problem gives us the coefficient of linear expansion of the material, and for the calculation of the volume
expansion, we need the coefficient of volumetric expansion
\[
\begin{gather}
\gamma =3\alpha \\
\gamma =3 \times 15 \times 10^{-6}\\
\gamma =45 \times 10^{-6}\;^{o}\text{C}^{-1}
\end{gather}
\]
The volume of the body, made of the same material as the container, after heating will be
\[ \bbox[#99CCFF,10px]
{\Delta V=\gamma V_{0}\Delta t}
\]
\[
\begin{gather}
V-V_{0}=\gamma V_{0}(\;t-t_{0}\;)\\
V=V_{0}+\gamma V_{0}(\;t-t_{0}\;)\\
V=V_{0}[\;1+\gamma (\;t-t_{0}\;)\;]\\
V=1 \times 10^{-3}[\;1+45 \times 10^{-6}(\;100-20\;)\;]\\
V=1 \times 10^{-3}[\;1+45 \times 10^{-6} \times 80\;]\\
V=1 \times 10^{-3}[\;1+3600 \times 10^{-6}\;]\\
V=1 \times 10^{-3}[\;1+0.0036\;]\\
V=1 \times 10^{-3} \times 1.0036
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{V=1.004 \times 10^{-3}\;\text{m}^{3}=1.004\;\text{l}}
\]