Problem 28 - Physics

Question

At a temperature of \(0^{\circ} \mathrm{C}\), the mass and volume of a fluid are \(825 \mathrm{~kg}\) and \(1.17 \mathrm{~m}^{3}\). The coefficient of volume expansion is \(1.26 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\) (a) What is the density of the fluid at this temperature? (b) What is the density of the fluid when the temperature has risen to \(20.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) Density at \(0^{\circ} \mathrm{C}\) is approximately \(705.13 \mathrm{~kg/m}^3\). (b) Density at \(20^{\circ} \mathrm{C}\) is approximately \(687.88 \mathrm{~kg/m}^3\).

Step 1: Understand Density Formula

Density is defined as mass per unit volume. The formula for density is \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.

Step 2: Calculate Initial Density

At \(0^{\circ} \mathrm{C}\), we have the mass \(m = 825 \mathrm{~kg}\) and the volume \(V = 1.17 \mathrm{~m}^3\). Substituting these values into the formula, the density is \( \rho = \frac{825}{1.17} \approx 705.13 \mathrm{~kg/m}^3 \).

Step 3: Apply Volume Expansion

The volume at a temperature change can be found using the formula for volume expansion: \( V' = V (1 + \beta \Delta T) \), where \( V' \) is the new volume, \( V \) is the original volume, \( \beta = 1.26 \times 10^{-3} \mathrm{C}^{-1} \) is the coefficient of volume expansion, and \( \Delta T = 20^{\circ} \mathrm{C}\) is the change in temperature.

Step 4: Calculate New Volume

Substitute into the formula: \( V' = 1.17 (1 + 1.26 \times 10^{-3} \times 20) = 1.17 (1 + 0.0252) = 1.17 \times 1.0252 \approx 1.1995 \mathrm{~m}^3 \).

Step 5: Calculate New Density

The mass of the liquid remains constant at \(825 \mathrm{~kg}\). Use the new volume to find the new density: \( \rho' = \frac{m}{V'} = \frac{825}{1.1995} \approx 687.88 \mathrm{~kg/m}^3 \).

Key Concepts

Density Formula

Density is a measure of how compact a substance is. It tells us how much mass is contained in a given volume. The density formula is given by \( \rho = \frac{m}{V} \), where:

• \( \rho \): Density of the substance
• \( m \): Mass of the substance
• \( V \): Volume of the substance

The formula shows the relationship between mass and volume; as mass increases with a constant volume, density increases, and vice-versa. It's crucial in many fields such as physics, engineering, and chemistry, to understand material properties.
In the exercise, the initial density was calculated by dividing the mass, \(825 \ \mathrm{kg}\), by the volume, \(1.17 \ \mathrm{m}^3\), which gives a density of approximately \(705.13 \ \mathrm{kg/m}^3\). Understanding this calculation is essential for evaluating changes when other variables like temperature affect volume.

Coefficient of Volume Expansion

The coefficient of volume expansion, denoted by \( \beta \), measures how much the volume of a material changes with temperature. It is defined as the fractional change in volume per degree of temperature change. The formula for calculating the change in volume due to thermal expansion is:

• \( V' = V (1 + \beta \Delta T) \)

Where:

• \( V' \): New volume after temperature change
• \( V \): Original volume
• \( \beta \): Coefficient of volume expansion
• \( \Delta T \): Change in temperature

This concept helps in predicting how substances expand when heated. In the given problem, the fluid's volume expands from \(1.17 \mathrm{~m}^3\) to approximately \(1.1995 \mathrm{~m}^3\) when temperature is raised by \(20^{\circ} \mathrm{C}\). It shows the practical use of the coefficient in real-world situations where temperature variations occur.

Temperature Change

Temperature change is a vital component in determining the physical properties of materials. It directly affects the volume, and consequently, the density, of substances. The relation to volume expansion is captured by the formula for volume expansion, where a higher temperature typically results in a larger volume.
In the exercise, the fluid experiences a temperature change from \(0^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\). This change in temperature results in the expansion of the fluid's volume due to the material's thermal properties. The final density was recalculated as \(687.88 \ \mathrm{kg/m}^3\) after the volume expanded to accommodate the increased temperature.

• Understanding how temperature impacts volume and density is crucial in many scenarios, including material selection and structural design, where thermal stability is a concern.