Chapter 13, Problem 9P

To determine

(a) To determine:

The expansion in a tabletop made of Super Invar for the given temperature rise

Answer to Problem 9P

Solution:

The expansion in the Super Invar tabletop is $2.16\times {10}^{-6}\text{m}$

Explanation of Solution

Given:

Coefficient of thermal expansion of Super Invar, ${\alpha }_{\text{S}\text{I}}=0.20\times {10}^{-6}/°C$

Original length of the tabletop, l₀ = 1.8 m

Rise in temperature, $\Delta \text{T}=6°C$

Formula used:

$\Delta l=l\times \alpha \times \Delta T,$ where α is the coefficient of linear expansion of a material.

Calculation:

The relation between change in length of a substance and change in temperature is given by,

$\Delta l=l\times \alpha \times \Delta T,$, where αT is the change in temperature andα is the coefficient of linear expansion of a material.

For the Super Invar tabletop,

$\Delta l_{SI}=\left(1.8m\right)\times (0.20\times {10}^{-6}/°C)\times \left(6°C\right)$$=2.16\times {10}^{-6}m$

To determine

(b) To determine:

The ratio of the expansion of a Super Invar tabletop to a steel tabletop

Answer to Problem 9P

Solution:

The ratio between the expansion in a Super Invar tabletop and a steel tabletop is 0.017

Explanation of Solution

Given:

Coefficient of thermal expansion of Super Invar, ${\alpha }_{SI}=0.20\times {10}^{-6}/°C$

Original length of the tabletop, $l_0=1.8m$

Rise in temperature, $\Delta T=6°C$

Formula used:

$\Delta l=l\times \alpha \times \Delta T$, where α is the coefficient of linear expansion of a material.

Calculation:

The relation between change in length of a substance and change in temperature is given by,

For a steel tabletop, $\alpha =12\times {10}^{-6}/°C$

Therefore, $\Delta l_S=\left(1.8m\right)\times (12\times {10}^{-6}/°C)\times \left(6°C\right)$

$=1.3\times {10}^{-4}m$

This gives us,

$\frac{\Delta l_{SI}}{\Delta l_S}$ = $\frac{2.16x{10}^{-6}m}{1.3x{10}^{-4}m}$ = 0.017