Answer:
(a) The length at temperature 180°C is 40.070 cm
(b) The length at temperature 90°C is 64.976 inches
(c) L(T, α) = 60·α·T - 9000·α + 60
Explanation:
(a) The given parameters are
The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C
The initial length of the steel L₀ = 40 cm
The initial temperature, t₀ = 40°C
The length at temperature 180°C = L
Therefore, from the given relation, for change in length, ΔL, we have;
ΔL = α × L₀ × ΔT
The amount the temperature changed ΔT = 180°C - 40°C = 140°C
Therefore, the change in length, ΔL, is found as follows;
ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm
Therefore, L = L₀ + ΔL = 40 + 0.07 = 40.07 cm
The length at temperature 180°C = 40.07 cm
(b) Given that the length at T = 120°C is 65 in., we have;
The temperature at which the new length is sought = 90°C
The amount the temperature changed ΔT = 90°C - 120°C = -30°C
ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches
The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.
The length at temperature 90°C = 64.976 inches
(c) L = L₀ + ΔL = L₀ + α × L₀ × ΔT = L₀ + α × L₀ × (T - T₀)
Therefore;
L = 60 + α × 60 × (T - 150°C)
L = 60 + α × 60 × T - 9000 × α
L(T, α) = 60·α·T - 9000·α + 60
= 0.668 mol Ni