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The change in volume of a quartz sphere with a diameter of 12.0 cm if its temperature is increased by 285° F is 0.214 cm³. From the complete version of the question, given the coefficient of volume expansion of squartz is [tex]1.50 \times 10^{-6} \textdegree C^{-1}[/tex].
The equation for the volume expansion is;
- ΔV = Vo β ΔT
We know that,
- Diameter of quartz = 12 cm
- Radius of quartz = 6 cm = 0.06 mm
- The coefficient of volume expansion of squartz (β) = [tex]1.50 \times 10^{-6} \textdegree C^{-1}[/tex]
- ΔT = 285° F
First, convert ΔT in fahrenheit to celsius.
- ΔT°F = 9/5 × ΔT°C
- ΔT°C = 5/9 × ΔT°F
- ΔT°C = 5/9 × 285°
- ΔT°C = 158.33°
Second, substitute all the value to the equation for the volume expansion.
- ΔV = Vo β ΔT
- ΔV = [tex]\frac{4}{3}.\frac{22}{7}.0.06^{3} \times 1.50 \times 10^{-6} \times 158.33[/tex]
- ΔV = [tex]2.14776 \times 10^{-7} m^3[/tex]
- ΔV = [tex]0.214 \times 10^{-6} m^{3}[/tex]
- ΔV = 0.214 cm³
Therefore, the change in volume of a quartz sphere with a diameter of 12.0 cm if its temperature is increased by 285° F is 0.214 cm³.
Learn more about the volume of gas and the temperature https://brainly.com/question/24519825
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