Question

a quartz sphere is 12.0 cm in diameter. what will be its change in volume if its temperature is increased by 285

Answer 1

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 $1.50 \times 10^{-6} \textdegree C^{-1}$.

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) = $1.50 \times 10^{-6} \textdegree C^{-1}$
• 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 =  $\frac{4}{3}.\frac{22}{7}.0.06^{3} \times 1.50 \times 10^{-6} \times 158.33$
• ΔV = $2.14776 \times 10^{-7} m^3$
• ΔV = $0.214 \times 10^{-6} m^{3}$
• 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.

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