Question

A rod with a length of 1.6 m is made of aluminum, which has a coefficient of linear expansion of 2.40 ✕ 10−5 (°C)−1. It is initially at a temperature of 25°C.

(a)
What is the change in length of the rod if the temperature drops to 0.0°C? (Give the absolute value in units of mm.)
mm
(b)
How does the rod's length change with lower temperature?
The rod shrinks with lower temperature.
The rod expands with lower temperature.
The rod's length does not change with lower temperature.
(c)
Find the absolute value of the fraction by which the rod's length changes between the two temperatures.

ΔL
L0

=

Answer 1

The aluminum rod's length will decrease by 0.96 mm when the temperature drops from 25°C to 0°C and this represents a fractional change of 0.0006 in the rod's length.

Step-by-step explanation:

The question involves calculating the change in length of an aluminum rod due to thermal contraction when the temperature decreases from 25°C to 0°C using the coefficient of linear expansion.

Part (a)

To find the change in length (absolute value in mm), we use the linear expansion formula: ΔL = αL₀ΔT. Here, α is the coefficient of linear expansion for aluminum (2.40 × 10⁵ (°C)⁻¹), L₀ is the initial length of the rod (1.6 meters), and ΔT is the change in temperature (-25°C).

ΔL = (2.40 × 10⁻µ)(1.6 m)(-25°C) = -0.00096 m or -0.96 mm.

The absolute value of the change in length is 0.96 mm.

Part (b)

The rod's length decreases or shrinks with lower temperature.

Part (c)

The absolute value of the fraction by which the rod's length changes between the two temperatures is given by |(ΔL)/L₀|.

|(ΔL)/L₀| = |(-0.96 mm)/(1600 mm)| = 0.0006

Answer 2

(a) To calculate the change in length of the rod, we can use the formula:

ΔL = αLΔT

where:

α = coefficient of linear expansion

L = initial length of the rod

ΔT = change in temperature

Substituting the given values, we get:

ΔL = (2.40 ✕ 10^-5 (°C)^-1) x (1.6 m) x (25°C - 0.0°C)

ΔL = 0.0096 m

Converting meters to millimeters, we get:

ΔL = 9.6 mm

Therefore, the change in length of the rod is 9.6 mm.

(b) The rod shrinks with lower temperature.

(c) The absolute value of the fraction by which the rod's length changes between the two temperatures can be calculated using the formula:

ΔL

L0

where:

ΔL = change in length of the rod

L0 = initial length of the rod

Substituting the given values, we get:

ΔL

L0

= 0.0096 m / 1.6 m

= 0.006

Therefore, the absolute value of the fraction by which the rod's length changes between the two temperatures is 0.006.