Question

A spherical steel ball bearing has a diameter of 2.540 cm at 27.00°C. (Assume the coefficient of linear expansion for steel is

11 ✕ 10⁻⁶ (°C)⁻¹.

(a) What is its diameter when its temperature is raised to 80.0°C? (Give your answer to at least four significant figures.)

(b) What temperature change is required to increase its volume by 0.800%?

Answer 1

Given that :

diameter of spherical steel ball, D = 0.0254 m

radius of spherical steel ball, r = 0.0127 m

initial temperature, T₀ = 27 ⁰C

final temperature, T_f = 80 ⁰C

coefficient of linear expansion for steel, _steel = 11 x 10^-60C^-1

(a) Using a volumetric thermal expansion, we have

V = V₀T

V = 3 V₀T

We know that,   V_f = V₀ + V = V₀ + 3 V₀T

V_f = V₀ (1 + 3T)

Therefore, we have

d_f = [V₀ + 3 V₀T]^1/3

d_f = (V₀)^1/3 [1 + 3T]^1/3

d_f = d₀ [1 + (1/3) (3T) + const*(3 V₀T)² + ....

d_f = d₀ [1 + T] = d₀ [1 + (T_f - T_i)]

d_f = (2.54 cm) [1 + (11 x 10^-60C^-1) (53 ⁰C)]

d_f = 2.541 cm

(b) What temperature change is required to increase its volume by 0.8%?

from an above formula, we get

V = 3 V₀T

(V / V₀) = 3T

(0.8 / 100) = 3 (11 x 10^-60C^-1) T

T = (0.008) / (33 x 10^-60C^-1)

T = 242.4 ⁰C