A cylinder of diameter 1.00000 cm at 30^∘ C is to be slid into a hole in a steel plate. The hol

step 1 Good thing. In this question, we will be solving for the final temperature of the plate. So give

A cylinder of diameter 1.00000 cm at 30^∘ C is to be slid...

A cylinder of diameter $1.00000 \mathrm{~cm}$ at $30^{\circ} \mathrm{C}$ is to be slid into a hole in a steel plate. The hole has a diameter of $0.99970 \mathrm{~cm}$ at 30 ${ }^{\circ} \mathrm{C}$. To what temperature must the plate be heated? For steel, $\alpha=$ $1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. We want the diameter of the hole to change by $$ \Delta L=(1.00000-0.99970) \mathrm{cm}=0.00030 \mathrm{~cm} $$ Using $\Delta L=\alpha L \Delta T$, $$ \Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C} $$ The temperature of the plate must be $30+27=57^{\circ} \mathrm{C}$.

A cylinder of diameter $1.00000 \mathrm{~cm}$ at $30^{\circ} \mathrm{C}$ is to be slid into a hole in a steel plate. The hole has a diameter of $0.99970 \mathrm{~cm}$ at 30 ${ }^{\circ} \mathrm{C}$. To what temperature must the plate be heated? For steel, $\alpha=$ $1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. We want the diameter of the hole to change by $$
\Delta L=(1.00000-0.99970) \mathrm{cm}=0.00030 \mathrm{~cm}
$$
Using $\Delta L=\alpha L \Delta T$,
$$
\Delta T=\frac{\Delta L}{\alpha L_{0}}=\frac{0.00030 \mathrm{~cm}}{\left(1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)(0.99970 \mathrm{~cm})}=27^{\circ} \mathrm{C}
$$
The temperature of the plate must be $30+27=57^{\circ} \mathrm{C}$.

Key Concepts

Thermal Expansion

Thermal expansion is the increase in the dimensions of a material as its temperature rises. It occurs because the particles within a material move more vigorously when heated, causing the material to occupy more space. This concept is fundamental in understanding how materials behave under temperature changes and plays a critical role in engineering and design applications.

Coefficient of Linear Expansion

The coefficient of linear expansion is a property of a material that measures the amount by which a unit length of the material expands per unit increase in temperature. It is typically denoted by the symbol ? and is expressed in inverse degrees Celsius (or Kelvin). This coefficient is essential in calculating the dimensional changes in engineering components when subjected to temperature variations.

Linear Expansion Formula

The linear expansion formula, ?L = ?L??T, relates the change in length (?L) of a material to its original length (L?), its coefficient of linear expansion (?), and the change in temperature (?T). This formula is widely used to predict how much a given material will expand or contract under thermal effects and is integral in designing components that must fit together under varying temperatures.

Dimensional Tolerances in Thermal Contexts

In engineering design, dimensional tolerances refer to the allowable limits of variation in a physical dimension. When thermal expansion is considered, these tolerances become crucial, as temperature changes can affect the fit and function of components. By understanding and applying thermal expansion principles, engineers can adjust temperatures or design parameters to ensure proper mechanical operation, even under fluctuating thermal conditions.

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