Answer:
[tex]8.1\cdot 10^{-4} C^{-1}[/tex]
Explanation:
The volumetric expansion of the liquid is given by
[tex]\Delta V=\alpha V_0 \Delta T[/tex]
where
[tex]\alpha[/tex] is the coefficient of volume expansion
[tex]V_0[/tex] is the initial volume
[tex]\Delta T[/tex] is the change in temperature
For the liquid in this problem,
[tex]V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3[/tex]
So we can solve the equation to find [tex]\alpha[/tex]:
[tex]\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}[/tex]
