ANSWER
[tex]\begin{gathered} (a)\text{ }285.76\text{ }mL \\ \\ (b)\text{ }284.67\text{ }mL \end{gathered}[/tex]EXPLANATION
Parameters given:
Linear expansivity of aluminum, α = 24 * 10^(-6) C^(-1)
Initial temperature,m T1 = 33°C
Initial volume of liquid, V1 = 285 mL
(a) To find the amount of liquid that the cup can hold on a summer day, apply the formula for volume expansivity:
[tex]3\alpha=\frac{V_2-V_1}{V_1(T_2-T_1)}[/tex]where T2 = temperature on a cold day
V2 = volume on a cold day
Therefore, solving for V2, we have that the volume that the cup can hold on a cold day will be:
[tex]\begin{gathered} 3*24*10^^{-6}=\frac{V_2-285}{285(-4-33)} \\ \\ 0.000072=\frac{V_2-285}{285*37}=\frac{V_2-285}{-10545} \\ \\ V_2-285=0.000072*-10545=0.75924 \\ \\ V_2=285+0.75924 \\ \\ V_2=285.76\text{ }mL \end{gathered}[/tex]That is the answer.
(b) To find the volume that the cup can hold on a spring day, apply the same formula but for T2 = 17°C:
[tex]\begin{gathered} 0.000072=\frac{V_2-285}{285(17-33)} \\ \\ 0.000072=\frac{V_2-285}{285*-16}=\frac{V_2-285}{-4560} \\ \\ V_2-285=0.000072*-4560=-0.32832 \\ \\ V_2=285-0.32832 \\ \\ V_2=284.67\text{ }mL \end{gathered}[/tex]That is the answer.
