Question

Calculate the increase in the volume of a liquid when heated to 40⁰C. Given that the original volume of liquid is 2 liters and its volume expansion coefficient is $12\times {10}^{-4}{/}^{0}C$.

Answer 1

Step 1:

Given data:

Increase in temperature of the liquid ($∆\mathrm{T}$) = 40⁰C

The original volume of liquid (V)= 2 liters

Volume expansion coefficient ($\mathrm{\gamma }$) of the liquid= $12\times {10}^{-4}{/}^{0}C$

Step 2:

Formula used:

$$\begin{gathered} \mathrm{Volume}\mathrm{expansion}\mathrm{coefficinet}\left(\mathrm{\gamma }\right)=\frac{\mathrm{increase}\mathrm{in}\mathrm{volume}(∆\mathrm{V})}{\mathrm{original}\mathrm{volume}\left(\mathrm{V}\right)\times \mathrm{increase}\mathrm{in}\mathrm{temperature}(∆\mathrm{T})} \\ \mathrm{That}\mathrm{is}, \\ \mathrm{\gamma }=\frac{∆\mathrm{V}}{\mathrm{V}\times ∆\mathrm{T}} \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{relation},\mathrm{we}\mathrm{get} \\ ∆\mathrm{V}=\mathrm{\gamma }\times \mathrm{V}\times ∆\mathrm{T} \end{gathered}$$

Step 3:

Calculation of the increase in the volume of the liquid:

$$\begin{gathered} ∆\mathrm{V}=\mathrm{\gamma }\times \mathrm{V}\times ∆\mathrm{T} \\ ∆\mathrm{V}=12\times {10}^{-4}\times 2\times 40 \\ ∆\mathrm{V}=24\times 40\times {10}^{-4} \\ ∆\mathrm{V}=960\times {10}^{-4}\mathrm{litres} \\ \mathrm{We}\mathrm{know}\mathrm{that}, \\ 1\mathrm{litre}=1000\mathrm{mL} \\ ={10}^3\mathrm{mL} \\ ∆\mathrm{V}=960\times {10}^{-4}\times {10}^3\mathrm{mL} \\ ∆\mathrm{V}=960\times {10}^{-1}\mathrm{mL} \\ ∆\mathrm{V}=96\mathrm{mL} \end{gathered}$$

Thus, the increase in the volume of the liquid will be 96 mL.