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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 \(20.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) Approximately 705.13 kg/m³ (b) Approximately 687.77 kg/m³.

Step by step solution

01

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:\[ \rho_0 = \frac{825 \text{ kg}}{1.17 \text{ m}^3} \approx 705.13 \text{ kg/m}^3. \]
02

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\[ \Delta V = 1.26 \times 10^{-3} \times 1.17 \times 20 \approx 0.029466 \; \text{m}^3. \]
03

Determine New Volume

The new volume \( V \) after temperature increase is the initial volume plus the change in volume: \[ V = V_0 + \Delta V = 1.17 \text{ m}^3 + 0.029466 \text{ m}^3 = 1.199466 \text{ m}^3. \]
04

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 \( V = 1.199466 \text{ m}^3 \).\[ \rho = \frac{825 \text{ kg}}{1.199466 \text{ m}^3} \approx 687.77 \text{ kg/m}^3. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{~kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

A \(0.35-\mathrm{kg}\) coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{~J} /\) \(\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{~kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C}\). To make a cup of coffeee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

A rock of mass 0.20 kg falls from rest from a height of \(15 \mathrm{~m}\) into a pail containing \(0.35 \mathrm{~kg}\) of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is \(1840 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). Ignore the heat absorbed by the pail itself, and determine the rise in the temperature of the rock and water.

Interactive Solution \(12.6112 .61\) at provides a model for solving problems such as this. A \(42-\mathrm{kg}\) block of ice at \(0{ }^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{~m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{~m} / \mathrm{s}\). Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at \(0{ }^{\circ} \mathrm{C}\).

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