Problem 97 - Physics

Question

An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. (a) In general, what factors determine how the volume of an object changes when it is heated? (b) Aluminum has a smaller coefficient of volume expansion than the liquid does. Which, if either, will expand more, the can or the liquid? Provide a reason for your answer. (c) How is the volume of liquid that spills over related to the changes in the volume of the can and the liquid? The can's initial volume at \(5^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-4} \mathrm{~m}^{3} .\) The coefficient of volume expansion for aluminum is \(69 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1}\), according to Table \(12-1\). When the can and the liquid are heated to \(78^{\circ} \mathrm{C}, 3.6 \times 10^{-6} \mathrm{~m}^{3}\) of liquid spills over. What is the coefficient of volume expansion of the liquid?

Step-by-Step Solution

Answer

The liquid has a larger volume expansion than the can, leading to the spillover. The coefficient of volume expansion for the liquid is approximately \( 130 \times 10^{-6} (^{\circ}C)^{-1} \).

Step 1: Understanding the Volume Expansion Concept

When an object is heated, its volume changes depending on its material properties. The key factor that determines how much the volume changes is the coefficient of volume expansion, denoted as \( \beta \). This coefficient describes how much a material expands per degree change in temperature.

Step 2: Comparing Expansion for Aluminum and Liquid

Given that aluminum has a smaller coefficient of volume expansion than the liquid, it indicates that for the same temperature change, the volume increase in the liquid will be greater than that of the aluminum can. Thus, the liquid will expand more than the aluminum can.

Step 3: Spill Over Volume Relationship

The volume of liquid that spills is the difference between the increase in the volume of the liquid and the aluminum can. Mathematically, it is expressed as \( V_{spill} = \Delta V_{liquid} - \Delta V_{can} \).

Step 4: Calculate the Volume Change in the Can

The volume change of the aluminum can due to temperature change \( \Delta T \) is calculated using \( \Delta V_{can} = \beta_{Aluminum} \times V_{initial} \times \Delta T \). With \( V_{initial} = 3.5 \times 10^{-4} \text{ m}^3 \), \( \Delta T = 78^{\circ}C - 5^{\circ}C = 73^{\circ}C \), and \( \beta_{Aluminum} = 69 \times 10^{-6} (^{\circ}C)^{-1} \), the volume change is \( \Delta V_{can} = 69 \times 10^{-6} \times 3.5 \times 10^{-4} \times 73 \).

Step 5: Calculate the Coefficient of Volume Expansion for the Liquid

Given the spillover volume \( 3.6 \times 10^{-6} \text{ m}^3 \), use the relationship from Step 3 to determine \( \beta_{liquid} \). Rearrange the expression to solve for \( \beta_{liquid} = \frac{\Delta V_{liquid}}{V_{initial} \times \Delta T} \), using the fact that \( \Delta V_{liquid} = \Delta V_{can} + V_{spill} \). Substitute \( V_{initial} = 3.5 \times 10^{-4} \text{ m}^3 \) and solve for \( \beta_{liquid} \).

Step 6: Solving for the Result

After calculating the aluminum's volume increase using \( \Delta V_{can} \), determine \( \Delta V_{liquid} \) with \( 3.5 \times 10^{-4} + 3.6 \times 10^{-6} \). Compute \( \beta_{liquid} = \frac{72.1 \times 10^{-6} + 3.6 \times 10^{-6}}{3.5 \times 10^{-4} \times 73} \).

Key Concepts

Coefficient of Volume Expansion

When materials are heated, they expand. The magnitude of this expansion depends on a property called the coefficient of volume expansion, represented as \( \beta \). This coefficient tells us how much the volume of a material changes for each degree of temperature change. It is usually expressed in units of \( (^{\circ}C)^{-1} \).

Each material has its unique coefficient. For example, aluminum has a lower \( \beta \) compared to many liquids, which means it expands less when heated by the same amount. Understanding \( \beta \) helps to predict how different materials will behave when subjected to temperature changes.

Temperature Change

Temperature change, denoted as \( \Delta T \), plays a crucial role in determining how much an object will expand. In the context of our exercise, the aluminum can and its liquid contents experience a temperature increase from \( 5^{\circ}C \) to \( 78^{\circ}C \).
This results in a \( \Delta T \) of \( 73^{\circ}C \).

The larger the temperature change, the greater the volume expansion, provided the material's \( \beta \) remains constant. Calculating the temperature difference is straightforward and is essential for using the expansion formula effectively.

Volume Expansion Calculation

To calculate how much an object expands, we use the formula: \( \Delta V = \beta \times V_{initial} \times \Delta T \).
This formula accounts for the object's initial volume \( V_{initial} \), its coefficient of volume expansion \( \beta \), and the temperature change \( \Delta T \).

In our scenario, the aluminum can's initial volume is \( 3.5 \times 10^{-4} \text{ m}^3 \) with \( \beta \) of \( 69 \times 10^{-6} (^{\circ}C)^{-1} \).
When plugged into the formula, you calculate the can's volume change as:\

• \( \Delta V_{can} = 69 \times 10^{-6} \times 3.5 \times 10^{-4} \times 73 \)

This computation is necessary to find how much extra volume is created when the material is heated.

Material Properties Effect on Volume

The material's intrinsic properties, particularly its \( \beta \), significantly influence how much it will expand. In this problem, the liquid inside the can has a higher \( \beta \) than aluminum, resulting in more significant expansion.
This is why more liquid spills over; its volume increases more than the can's.

The spilled volume is calculated by finding the difference between the expansion of the liquid and the aluminum: \( V_{spill} = \Delta V_{liquid} - \Delta V_{can} \).

• This formula shows that more expansion from a higher \( \beta \) results in overflow when the container expansion is less.

Understanding how material properties affect volume is key to predicting behaviors in thermal management.