Question

At room temperature, an iron horseshoe, when dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm})\) causes the water level to rise \(0.25 \mathrm{~cm}\) above the level without the horseshoe in the tank. When heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K}\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

Short answer

Answer: To find the new rise in water level, follow these steps: 1. Calculate the initial volume of the horseshoe at room temperature using the given water level rise. 2. Determine the volume expansion of the horseshoe due to the increase in temperature using the linear expansion coefficient. 3. Find the final volume of the heated horseshoe by adding the volume expansion to the initial volume. 4. Compute the new rise in water level by dividing the final volume by the base area of the tank (πR²). By following these steps, you will find the new rise in water level when the heated horseshoe is dunked into the cylindrical tank.

Step-by-step solution

Calculate the initial volume of the horseshoe

First, we need to find out the initial volume of the horseshoe at room temperature. To do so, we can find the volume of water corresponding to the rise in water level when the horseshoe is dunked into the tank: \(V_{water} = \pi R^2 h_{rise}\) where \(R = 10.0 \mathrm{~cm}\) is the radius of the cylindrical tank, and \(h_{rise} = 0.25 \mathrm{~cm}\) is the rise in water level. Plug in the values to find the initial volume of the horseshoe: \(V_{1} = V_{water} = \pi (10.0)^2 (0.25)\)

Calculate the expansion in the horseshoe's volume

Next, we need to find how much the horseshoe has expanded due to the increase in temperature. The linear expansion coefficient (\(\alpha\)) is given as \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\). The initial temperature of the horseshoe is room temperature, which is \(20^{\circ}\mathrm{C}\). The final temperature is \(7.00\cdot 10^{2} \mathrm{~K}\). Since the temperatures are in different units, we need to convert the final temperature to Celsius: \(T_{2} = 7.00\cdot 10^{2} \mathrm{~K} - 273.15 = 426.85^{\circ}\mathrm{C}\) Now, we can compute the change in temperature: \(\Delta T = T_{2} - T_{1} = 426.85 - 20 = 406.85^{\circ}\mathrm{C}\) Since the volume expansion for a solid is given by the following formula: \(\Delta V = V_{1} \cdot 3 \alpha \Delta T\) Plug in the values to find the volume expansion: \(\Delta V = V_{1} \cdot 3 \cdot (11 \cdot 10^{-6}) \cdot 406.85\)

Calculate the final volume of the horseshoe

Now that we've determined the volume expansion due to the increase in temperature, we can find the expanded volume of the horseshoe: \(V_{2} = V_{1} + \Delta V\)

Determine the new rise in water level

Having found the final volume of the heated horseshoe, we can now determine the new rise in water level: \(h_{new} = \frac{V_{2}}{\pi R^2}\) Finally, we will have the rise in water level when the heated horseshoe is dunked back into the cylindrical tank.

Key concepts

Volume Calculation

In physics, calculating volume is a crucial skill, especially when dealing with submerged objects. Imagine dunking an object into a tank of water. The water level rises due to the object's volume displacing the water.
To find this volume, use the formula for the volume of water displaced. For our problem, this is

• \(V_{water} = \pi R^2 h_{rise}\)

where:

• \(R\) is the radius of the container, which is given as \(10.0 \text{ cm}\), and
• \(h_{rise}\) is the height the water rises, 0.25 cm.

When the horseshoe is dunked into the tank, it displaces enough water to raise the water level by 0.25 cm. By calculating the volume of this displaced water, we determine the horseshoe's volume at room temperature.
Using the given values, substitute them into the equation to find the initial volume \(V_1\) of the horseshoe.

Linear Expansion Coefficient

When objects are heated, they expand unless constrained. This expansion can be quantified using the linear expansion coefficient. For iron, which in this context is similar to steel, the linear expansion coefficient \(\alpha\) is about \(11 \cdot 10^{-6} { }^{\circ} \mathrm{C}^{-1}\).
This coefficient indicates how much the material will change in length per degree change in temperature.
If a material expands linearly, understanding this coefficient helps predict how much it expands when heated. For volumetric expansion, the relationship involves the initial volume \(V_1\), the expansion coefficient, and the temperature change \(\Delta T\). The formula for volume change is:

• \(\Delta V = V_1 \cdot 3 \cdot \alpha \cdot \Delta T\)

Here, the factor of 3 accounts for the three dimensions of volume (length, width, height). As the temperature increases, the product of these terms will provide the increase in volume.

Temperature Conversion

In physics, it is common to convert temperature units, as different scientific equations require specific units. The most common conversions are between Kelvin and Celsius. For instance, room temperature in Celsius (about \(20{ }^{\circ} \mathrm{C}\)) is simple to convert to Kelvin by adding 273.15.
However, in our problem, the iron horseshoe is heated to \(700 \mathrm{~K}\).
To convert this back to Celsius (which many equations prefer), subtract 273.15. Performing this conversion gives:

• Final temperature \(T_2 = 700 - 273.15 = 426.85^{\circ}\text{C}\)

Using this Celsius temperature helps in computing the temperature difference \(\Delta T\), which is essential in calculating expansion.

Physics Problem Solving

Solving physics problems involves a structured step-by-step approach to understanding the involved quantities and relationships. Start by identifying known values, like the initial volume using displacement principles. Recognize given constants, such as coefficients of expansion, and identify the temperature changes.
These enable calculations of volume changes and final volume using:

• \(V_{2} = V_{1} + \Delta V\)

Once you've determined the expanded volume, revisit the displacement principle to find the new rise in water level with

• \(h_{new} = \frac{V_{2}}{\pi R^2}\)

By following these logical steps, you can break down complex problems into simpler, manageable parts. This not only helps in arriving at a solution but also deepens your understanding of the physics involved.