Question

When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coefficient of volume expansion? (The answer is ________ •10^-4 C^-1. Just fill in the blank.)

Answer 1

Answer:

$8.1\cdot 10^{-4} C^{-1}$

Explanation:

The volumetric expansion of the liquid is given by

$\Delta V=\alpha V_0 \Delta T$

where

$\alpha$ is the coefficient of volume expansion

$V_0$ is the initial volume

$\Delta T$ is the change in temperature

For the liquid in this problem,

$V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3$

So we can solve the equation to find $\alpha$:

$\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}$