Answer:
a) 2.00024 m
b) 0.036%
c) 436.67°C
Explanation:
Given
Initial length = L₀ = 2 m
Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²
We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³
Initial Temperature = T₀ = 20°C
Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹
a) New length of the rod after heating to 80°C
Linear expansion is given as
ΔL = L₀ × α ×ΔT
ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm
New length = old length + expansion = 2 + 0.00024 = 2.00024 m
b) The percentage of the volume change of the rod.
Volume expansion is given by
ΔV = V₀ × (3α) × ΔT
Volume expansivity ≈ 3 × (linear expansivity)
ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³
Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%
c) The maximal temperature we can allow if the volume should not increase by more than half percent.
For a half percent increase in volume, the corresponding change in volume needs to be first calculated.
Percentage change in volume = 100% × (ΔV/V₀)
0.5 = 100% × (ΔV/0.04)
(ΔV/0.04) = 0.005
ΔV = 0.0002 m³
Then we now investigate the corresponding temperature that causes this.
ΔV = V₀ × (3α) × ΔT
0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT
ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C
Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C
is the potential difference applied to it.
