A concrete block with a volume of [tex]2.00 \, \text{m}^3[/tex] is heated from [tex]22.5^\circ \text{C}[/tex] to [tex]627^\circ \text{C}[/tex]. By how many [tex]\text{m}^3[/tex] will the volume of the block increase?

Use the coefficient of volume expansion from our class textbook. Give your answer to 3 significant digits. Do not include the units with the answer.



Answer :

Certainly! Let's solve the problem step-by-step.

1. Identify the initial conditions and constants:
- Initial volume ([tex]\( V_i \)[/tex]): 2.00 m³
- Initial temperature ([tex]\( T_i \)[/tex]): 22.5°C
- Final temperature ([tex]\( T_f \)[/tex]): 627°C

2. Determine the coefficient of volume expansion for concrete:
- From the class textbook, the coefficient of volume expansion ([tex]\( \beta \)[/tex]) is [tex]\( 3.6 \times 10^{-5} \)[/tex] [tex]\( \text{°C}^{-1} \)[/tex].

3. Calculate the change in temperature:
[tex]\[ \Delta T = T_f - T_i = 627^\circ\text{C} - 22.5^\circ\text{C} = 604.5^\circ\text{C} \][/tex]

4. Calculate the increase in volume ([tex]\( \Delta V \)[/tex]):
[tex]\[ \Delta V = V_i \times \beta \times \Delta T \][/tex]
Plugging in the values:
[tex]\[ \Delta V = 2.00 \, \text{m}^3 \times 3.6 \times 10^{-5} \, \text{°C}^{-1} \times 604.5\,^\circ\text{C} \][/tex]

5. Perform the multiplication to find the increase in volume:
[tex]\[ \Delta V = 2.00 \times 3.6 \times 10^{-5} \times 604.5 \][/tex]
[tex]\[ \Delta V = 0.043536 \][/tex]

6. Round the answer to 3 significant digits:
[tex]\[ \Delta V \approx 0.044 \][/tex]

So, the volume of the concrete block will increase by approximately 0.044 m³.