Answer:
[tex]\Delta V=1.667\times10^{-3}[/tex]
Explanation:
Given the initial temperature T_i=2° C
final temperature T_f= 32° C
The original volume of water Vo=268.8 mL= 0.2688 L
we need to calculate the change in the volume
As we know that volume expansion is given by
[tex]\frac{\Delta V}{V_0}= \beta\Delta T[/tex]
ΔV= change in Volume
β= expansion coefficient = [tex]207\times10^{-6} K^{-1}[/tex]
therefore,
[tex]\Delta V= \beta\Delta T V_0[/tex]
plugging values we get
[tex]\Delta V=207\times10^{-6} K^{-1} (32-2)\times0.2688[/tex]
[tex]\Delta V=1.667\times10^{-3}[/tex]
