Answer:
The original volume of the metal sphere is approximately 1 cm³
Step-by-step explanation:
The given parameters are;
The temperature change of the metal ball, ΔT = 30°C = 30 K
The new volume of the metal ball, V₂ = 1.0018 cm³
The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹
We have;
d₂ = d₁·(1 + α·ΔT)
Where;
d₁ = The original diameter of the metal ball
d₂ = The new diameter of the ball
From the volume of the ball, V₂, we have;
V₂ = 1.0018 cm³ = (4/3)×π×r₂³
Where;
r₂ = The new radius = d₂/2
∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³
∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm
d₁ = d₂/(1 + α·ΔT)
∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm
The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³
∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³
The original volume of the metal sphere, V₁ ≈ 1 cm³.