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Chapter 19: Problem 51

A liquid has a density \(\rho\). (a) Show that the fractional change in density for a change in temperature \(\Delta T\) is \(\Delta \rho / \rho=-\beta \Delta T\). What does the negative sign signify? (b) Fresh water has a maximum density of \(1.0000 \mathrm{g} / \mathrm{cm}^{3}\) at \(4.0^{\circ} \mathrm{C} .\) At \(10.0^{\circ} \mathrm{C},\) its density is \(0.9997 \mathrm{g} / \mathrm{cm}^{3} .\) What is \(\beta\) for water over this temperature interval?

Short Answer

Expert verified
The negative sign indicates that as the temperature increases, the density generally decreases. 尾 for water over the given temperature interval is +0.00005 掳C^-1.

Step by step solution

01

Solution for Part (a)

The change in volume (鈭哣), due to temperature change (鈭員) is given as 鈭哣 = 尾V鈭員. The volume is inversely proportional to density for a given mass. Therefore, volume V = m/蟻, where m is the mass and 蟻 is the density. So, 鈭哣 = 鈭�(m/蟻) = m(1/蟻 * 鈭喯�). As a result, we obtain m/蟻 * 鈭喯� = 尾m/蟻鈭員. Dividing m/蟻 from both sides, we get 鈭喯�/蟻 = -尾鈭員. The negative sign indicates that as the temperature T increases, the density 蟻 usually decreases.
02

Solution for Part (b)

Given: initial density 蟻1 = 1.0000 g/cm^3 at temperature T1 = 4.0掳C, final density 蟻2 = 0.9997 g/cm^3 at temperature T2 = 10.0掳C. It's important to note we are interested in the absolute change. Therefore, 鈭員 = T2 - T1 = 10.0掳C - 4.0掳C = 6.0掳C, and 鈭喯�/蟻 = 蟻2/蟻1 - 1 = 0.9997/1.0000 - 1 = -0.0003. By substituting 鈭喯�/蟻 and 鈭員 in (鈭喯�/蟻 = -尾鈭員), and solving for 尾, we find 尾 = - 鈭喯�/ (蟻 * 鈭員) = - (-0.0003)/ (6.0) = +0.00005 掳C^-1.

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Key Concepts

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

Thermal Expansion
Objects and materials generally expand when heated, a phenomenon known as thermal expansion. This is a common characteristic of most substances and results in a change in dimensions, such as an increase in length, area, or volume.
  • In solids, thermal expansion is mainly about changes in length, called linear expansion.
  • In liquids, it becomes more about changes in volume, as they spread out when heated.
For liquids, thermal expansion causes the density to decrease because the same mass occupies a larger volume. This is why the density of liquids like water changes with temperature.
Volume Change
When considering volume change due to temperature, it鈥檚 vital to comprehend how the volume of a material varies as it heats up or cools down. For a liquid, we can describe the change in volume (\(\Delta V\)) caused by a change in temperature (\(\Delta T\)) using the equation:\[\Delta V = \beta V \Delta T\]Here, \(\beta\) is the coefficient of volumetric expansion, and \(V\) is the original volume.Volume change impacts density, which is mass divided by volume. Because the mass remains constant, as the volume of a liquid increases, its density decreases.
Temperature Dependence
Temperature dependence is crucial for understanding how materials react to thermal fluctuations. The properties of substances, such as density, are temperature-dependent. As temperature rises, liquid molecules gain energy, causing them to move further apart and decreasing density. This temperature-dependent behavior is predictable and quantifiable, allowing scientists to calculate changes in material properties based on temperature changes. In practice, water is a unique case. Its density reaches a maximum at around 4掳C, and any temperature increase or decrease from this point leads to a decrease in density.
Coefficient of Volumetric Expansion
The coefficient of volumetric expansion (\(\beta\)) is a material-specific value that quantifies the rate of volume change per degree change in temperature. It provides insight into how much a material will expand or contract with temperature variations.
  • For water, the coefficient is relatively small, reflecting its moderate response to temperature changes.
  • The value is determined from experimental data and varies with pressure and temperature conditions.
In our example with water, we calculated \(\beta\) using the formula:\[\beta = \frac{-\Delta \rho}{\rho \Delta T}\]This exemplifies how science allows us to detail the thermal behavior of liquids, facilitating further applications like designing cooling systems or understanding climate dynamics.

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Most popular questions from this chapter

One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of \(27.0^{\circ} \mathrm{C} .\) (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated until both the pressure and volume are doubled, what is the final temperature?

A diving bell in the shape of a cylinder with a height of \(2.50 \mathrm{m}\) is closed at the upper end and open at the lower end. The bell is lowered from air into sea water \(\left(\rho=1.025 \mathrm{g} / \mathrm{cm}^{3}\right) .\) The air in the bell is initially at \(20.0^{\circ} \mathrm{C} .\) The bell is lowered to a depth (measured to the bottom of the bell) of 45.0 fathoms or \(82.3 \mathrm{m}\). At this depth the water temperature is \(4.0^{\circ} \mathrm{C},\) and the bell is in thermal equilibrium with the water. (a) How high does sea water rise in the bell? (b) To what minimum pressure must the air in the bell be raised to expel the water that entered?

A brass ring of diameter \(10.00 \mathrm{cm}\) at \(20.0^{\circ} \mathrm{C}\) is heated and slipped over an aluminum rod of diameter \(10.01 \mathrm{cm}\) at \(20.0^{\circ} \mathrm{C} .\) Assuming the average coefficients of linear expansion are constant, (a) to what temperature must this combination be cooled to separate them? Is this attainable? (b) What If? What if the aluminum rod were \(10.02 \mathrm{cm}\) in diameter?

A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a temperature of \(-80.0^{\circ} \mathrm{C}\) ) and in boiling ethyl alcohol \(\left(78.0^{\circ} \mathrm{C}\right) .\) The two pressures are \(0.900 \mathrm{atm}\) and \(1.635 \mathrm{atm} .\) (a) What Celsius value of absolute zero does the calibration yield? What is the pressure at (b) the freezing point of water and (c) the boiling point of water?

A volumetric flask made of Pyrex is calibrated at \(20.0^{\circ} \mathrm{C} .\) It is filled to the 100 -mL mark with \(35.0^{\circ} \mathrm{C}\) acetone. (a) What is the volume of the acetone when it cools to \(20.0^{\circ} \mathrm{C} ?\) (b) How significant is the change in volume of the flask?
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