$147.5 \mathrm{ml}$ of ethyl alcohol (coefficient of volume expansion $\beta=1,120 \times 10^{-6} \mathrm{~K}^{-1}$ ), at a temperature of 273.1 $\mathrm{K}$, is poured into a $150.0 \mathrm{ml}$ glass beaker (whose coefficient of volume expansion is negligible by comparison). To what temperature does the beaker have to be warmed for it to be completely full (i.e., for the volume of the alcohol to reach $150.0 \mathrm{ml})$ ?

A) $288.0 \mathrm{~K}$
B) $283.1 \mathrm{~K}$
C) $278.1 \mathrm{~K}$
D) $288.2 \mathrm{~K}$

Question

147.5 \mathrm{ml}

of ethyl alcohol (coefficient of volume expansion

\beta=1,120 \times 10^{-6} \mathrm{~K}^{-1}

) , at a temperature of 273.1

\mathrm{K}

, is poured into a

150.0 \mathrm{ml}

glass beaker (whose coefficient of volume expansion is negligible by comparison) . To what temperature does the beaker have to be warmed for it to be completely full (i.e., for the volume of the alcohol to reach

150.0 \mathrm{ml})

?

A) $288.0 \mathrm{~K}$
B) $283.1 \mathrm{~K}$
C) $278.1 \mathrm{~K}$
D) $288.2 \mathrm{~K}$

Quizplus · Accepted Answer

The volume expansion of the alcohol can be described by the equation $\Delta V = V_0 \beta \Delta T$, where $\Delta V$ is the change in volume, $V_0$ is the initial volume, $\beta$ is the coefficient of volume expansion, and $\Delta T$ is the change in temperature. We want the final volume $V_f = 150.0 \, \text{ml}$, so $\Delta V = V_f - V_0 = 150.0 \, \text{ml} - 147.5 \, \text{ml} = 2.5 \, \text{ml}$. Plugging in the values, we get $2.5 = 147.5 \times 1,120 \times 10^{-6} \times \Delta T$. Solving for $\Delta T$, we find $\Delta T = \frac{2.5}{147.5 \times 1,120 \times 10^{-6}} \approx 15.1 \, \text{K}$. Adding this to the initial temperature of $273.1 \, \text{K}$, we get a final temperature of approximately $288.2 \, \text{K}$.

147.5ml147.5 \mathrm{ml}147.5ml Of Ethyl Alcohol (Coefficient of Volume Expansion β=1,120×10−6 K−1\beta=1,120 \times 10^{-6} \mathrm{~K}^{-1}β=1,120×10−6 K−1

$147.5 \mathrm{ml}$ Of Ethyl Alcohol (Coefficient of Volume Expansion $\beta=1,120 \times 10^{-6} \mathrm{~K}^{-1}$