Chapter 40, Problem 154A

To determine

(a)

The value of expansion of the aluminum.

Answer to Problem 154A

The value of expansion of the aluminum is $1293.278{\text{×10}}^{\text{-5}}\text{in}$.

Explanation of Solution

Given:

Original length of metal = $\text{6}\text{.7520in}\text{.}$

Linear expansion per unit of length per degree Fahrenheit = $\text{ 1}{\text{.22×10}}^{\text{-5}}$

Original temperature = $\text{68}\text{.0°F}$

Temperature to which heated = $\text{225}\text{.0°F}$

According to the given formula of expansion,

  $\text{Expansion = original}\text{length}\times \text{Linear expansion per unit}\text{length}\text{per}\text{degree Fahrenheit }\times \Delta \text{T}$

On substituting the given values,

  $\text{Tempreature change}\left(\Delta \text{T}\right)\text{=}\text{225}\text{.0°F}-6\text{8}\text{.0°F}=157\text{°F}$

  $\begin{array}{l}\text{Expansion = }\left(\text{6}\text{.7520in}\right)\times \left(\text{1}{\text{.22×10}}^{\text{-5}}\right)\times 157\text{°F}\\ \text{Expansion = }1293.278{\text{×10}}^{\text{-5}}\text{in}\end{array}$

So, the value of expansion of the aluminum is $1293.278{\text{×10}}^{\text{-5}}\text{in}$.

To determine

(b)

The value of expansion of the copper.

Answer to Problem 154A

The value of expansion of the copper is $19948.5{\text{×10}}^{\text{-6}}\text{ft}\text{.}$

Explanation of Solution

Given:

Original length of metal = $\text{35}\text{.750ft}$

Linear expansion per unit of length per degree Fahrenheit = $\text{9}{\text{.000×10}}^{\text{-6}}$

Original temperature = $\text{35}\text{.0°F}$

Temperature to which heated = $\text{97}\text{.0°F}$

According to the given formula of expansion,

  $\text{Expansion = original}\text{length}\times \text{Linear expansion per unit}\text{length}\text{per}\text{degree Fahrenheit }\times \Delta \text{T}$

On substituting the values,

  $\text{Tempreature change}\left(\Delta \text{T}\right)\text{=}\text{97}\text{.0°F}-35.\text{0°F}=62.0\text{°F}$

  $\begin{array}{l}\text{Expansion = }\left(\text{35}\text{.750ft}\right)\times \left(\text{9}{\text{.000×10}}^{\text{-6}}\right)\times \left(\text{62}\text{.0°F}\right)\\ \text{Expansion = }19948.5{\text{×10}}^{\text{-6}}\text{ft}\text{.}\end{array}$

So, the value of expansion of the copper is $19948.5{\text{×10}}^{\text{-6}}\text{ft}\text{.}$

To determine

(c)

The value of expansion of the carbon steel.

Answer to Problem 154A

The value of expansion of the carbon steel is $125540.390{\text{×10}}^{\text{-6}}\text{in}$.

Explanation of Solution

Given:

Original length of metal = $\text{30}\text{.0950in}\text{.}$

Linear expansion per unit length per degree Fahrenheit = $\text{6}{\text{.330×10}}^{\text{-6}}$

Original temperature = $\text{84}\text{.0°F}$

Temperature to which heated = $\text{743}\text{.0°F}$

According to the given formula of expansion,

  $\text{Expansion = original}\text{length}\times \text{Linear expansion per unit}\text{length}\text{per}\text{degree Fahrenheit }\times \Delta \text{T}$

On substituting the values,

  $\text{Tempreature change}\left(\Delta \text{T}\right)\text{=}\text{743}\text{.0°F}-84.\text{0°F}=659.0\text{°F}$

  $\begin{array}{l}\text{Expansion = }\left(\text{30}\text{.0950in}\right)\times \left(\text{6}{\text{.330×10}}^{\text{-6}}\right)\times 659.0\text{°F}\\ \text{Expansion = }125540.3897{\text{×10}}^{\text{-6}}\text{in}=125540.390{\text{×10}}^{\text{-6}}\text{in}\end{array}$

So, the value of expansion of the carbon steel is $125540.390{\text{×10}}^{\text{-6}}\text{in}$.