Question

A steel scale measures the length of a copper rod as 80cm when both are at 20°c , the caliberation temperature of the scale . what would the scale read for the length of the rod when both are at 40°c,

given coefficient of linear expansion of steel =1.1×10^-5°c^-1
coefficient of linear expansion of copper =1.7×10^-5°c^-1
a. 80.0096 cm
b. 80.0272 cm
c. 1 cm
d. 25.2 cm

Answer 1

a) 80.0096 cm

Step-by-step explanation:

This is an exercise of thermal expansion, we must calculate how much each material is dilated and the difference is the expansion of the measurement

The expression for linear dilation

ΔL = α L₀ ΔT

Let's start with the steel

ΔL / L₀ = 1.1 10⁻⁵ (40 -20)

ΔL / L₀ = 22 10⁻⁵

Copper

ΔL / L₀ = 1.7 10⁻⁵ (40-20)

ΔL / L₀ = 34.0 10⁻⁵

What made the copper bar high is the differentiated being read

ΔD = ΔL / L₀_copper - ΔL / L₀_Steel

The initial length of the steel balance is made equal to the length of the copper rod

ΔL = (34 - 22) 10⁻⁵ / 80

ΔL = 0.0096

For which the final length of the copper bar

`L_(f)` = L₀ +ΔL

`L_(f)` = 80 +0.0096

`L_(f)` = 80.0096 cm

The correct answer is a