Temperature 21973

The aluminum container has an internal volume of 0.75 l at 20 °C. How does this volume change if the temperature increases by 55 °C? A = 24.10^-6 1/K.

Correct answer:

V₂ =  0.753 l

Step-by-step explanation:

$$\begin{gathered} t_1=20\text{°C} \\ {V}_1=0.75\text{l} \\ T=55\text{°C} \\ {t}_2=t_1+55=20+55=75\text{°C} \\ \alpha =24\cdot 10^{-6}\doteq 2.4\cdot 10^{-5}1\text{/K} \\ {V}_2=V_1\cdot (1+3\cdot \alpha \cdot T)=0.75\cdot (1+3\cdot 2.4\cdot 10^{-5}\cdot 55)=0.753\text{l} \end{gathered}$$

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