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) Approximately 705.13 kg/m³ (b) Approximately 687.77 kg/m³.

Step 1: Find the Initial Density

The formula for density \( \rho \) is given by \( \rho = \frac{m}{V} \) where \( m \) is the mass and \( V \) is the volume. Initially, the mass is given as 825 kg and the volume is 1.17 m³. Calculating the initial density, we have:\[ \rho_0 = \frac{825 \text{ kg}}{1.17 \text{ m}^3} \approx 705.13 \text{ kg/m}^3. \]

Step 2: Calculate Volume Change

Use the formula for the change in volume due to temperature change: \( \Delta V = \beta V_0 \Delta T \). Here, \( \beta = 1.26 \times 10^{-3} \; (\mathrm{C}^\circ)^{-1} \), \( V_0 = 1.17 \; \text{m}^3 \), and \( \Delta T = 20^\circ \text{C} \). Then\[ \Delta V = 1.26 \times 10^{-3} \times 1.17 \times 20 \approx 0.029466 \; \text{m}^3. \]

Step 3: Determine New Volume

The new volume \( V \) after temperature increase is the initial volume plus the change in volume: \[ V = V_0 + \Delta V = 1.17 \text{ m}^3 + 0.029466 \text{ m}^3 = 1.199466 \text{ m}^3. \]

Step 4: Calculate New Density

The density \( \rho \) at the new temperature can be found using \( \rho = \frac{m}{V} \), where \( m = 825 \text{ kg} \) and the new volume \( V = 1.199466 \text{ m}^3 \).\[ \rho = \frac{825 \text{ kg}}{1.199466 \text{ m}^3} \approx 687.77 \text{ kg/m}^3. \]

Key Concepts

Coefficient of Volume Expansion

The coefficient of volume expansion, often denoted as \( \beta \), is a measure of how much a substance's volume changes in response to a change in temperature. This property is particularly important for fluids, as they often experience significant volume changes with temperature. Understanding \( \beta \) helps in predicting how a fluid will expand or contract when heated or cooled.
To calculate the change in volume due to temperature change, we use the formula:

• \( \Delta V = \beta V_0 \Delta T \)

Here, \( \Delta V \) represents the change in volume, \( V_0 \) is the initial volume, and \( \Delta T \) is the change in temperature. The unit of \( \beta \) is typically \( (\text{C}^\circ)^{-1} \), indicating how much the volume changes per degree Celsius temperature change.
This concept is crucial in various fields like engineering and science, particularly when dealing with systems where temperature fluctuations can affect performance or safety.

Temperature Effects on Volume

Temperature can profoundly impact the volume of a fluid, leading to either expansion or contraction. When a fluid is heated, its particles move more vigorously, causing the substance to expand and occupy more space. Conversely, cooling the fluid will typically cause it to contract, as the kinetic energy of its particles decreases. This principle is observed everywhere in our daily lives, such as in thermometers or hot air balloons.
The relationship between the temperature change and volume change in a fluid is primarily governed by its coefficient of volume expansion, \( \beta \). As temperature increases, the new volume \( V \) can be computed using the formula:

• \( V = V_0 + \Delta V \)

Where \( \Delta V = \beta V_0 \Delta T \). This understanding is essential for predicting behaviors in various applications, from environmental science to industrial processes.

Fluid Density Calculation

Fluid density signifies how concentrated a fluid's mass is in a given volume and is symbolized by \( \rho \). It is calculated using the formula:

• \( \rho = \frac{m}{V} \)

where \( m \) is the mass of the fluid, and \( V \) is its volume. Density is an essential property influencing buoyancy, pressure, and even the mixing of fluids.
In scenarios where temperature affects the fluid's volume, as demonstrated in the original exercise, recalculating density after temperature-induced volume change shows how density decreases with volume expansion. Thus, while mass remains constant, any increase in volume due to rising temperature results in a decrease in density:

• Initial Density: \( \rho_0 = \frac{m}{V_0} \)
• New Density After Volume Change: \( \rho = \frac{m}{V} \)

This decrease can have significant implications in fluid dynamics and thermodynamics, affecting how fluids move and react under different conditions.