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Chapter 12: Problem 36

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. 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 \(69 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) 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?

Short Answer

Expert verified
The coefficient of volume expansion of the liquid is approximately \(8.29 \times 10^{-5} \text{C}^{-1}\).

Step by step solution

01

Determine Temperature Change

Calculate the change in temperature (螖T) experienced by the can and the liquid. The final temperature is 78掳C and the initial temperature is 5掳C. Therefore, \[ \Delta T = 78^{\circ} \text{C} - 5^{\circ} \text{C} = 73^{\circ} \text{C}. \]
02

Calculate Volume Expansion of Can

Use the formula for volume expansion, which is \( \Delta V = \beta V_0 \Delta T \), where \( \beta \) is the coefficient of volume expansion, \( V_0 \) is the initial volume, and \( \Delta T \) is the temperature change.For the can, \( \beta = 69 \times 10^{-6} \text{C}^{-1} \) and \( V_0 = 3.5 \times 10^{-4} \text{m}^3 \). Therefore, \[ \Delta V_{\text{can}} = (69 \times 10^{-6}) \times (3.5 \times 10^{-4}) \times 73 = 1.76415 \times 10^{-5} \text{m}^3. \]
03

Determine Liquid Spillage

The spillage amount is given as \( 3.6 \times 10^{-6} \text{m}^3 \). This is the extra volume increase of the liquid beyond the can's expansion, indicating how much more the liquid has expanded compared to the can.
04

Calculate Liquid's Total Expansion

The total expansion of the liquid is the sum of the volume expansion of the can and the spillage:\[ \Delta V_{\text{liquid}} = \Delta V_{\text{can}} + \text{spillage} = 1.76415 \times 10^{-5} + 3.6 \times 10^{-6} = 2.12415 \times 10^{-5} \text{m}^3. \]
05

Calculate Coefficient of Volume Expansion for Liquid

Now solve for the coefficient of volume expansion for the liquid using \( \Delta V = \beta V_0 \Delta T \).Rearrange to find \( \beta_{\text{liquid}} \):\[ \beta_{\text{liquid}} = \frac{\Delta V_{\text{liquid}}}{V_0 \Delta T} = \frac{2.12415 \times 10^{-5}}{3.5 \times 10^{-4} \times 73}. \]Calculate:\[ \beta_{\text{liquid}} \approx 8.29 \times 10^{-5} \text{C}^{-1}. \]

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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 is a measure of how much a material's volume changes when it is heated or cooled. It is denoted by the Greek letter \( \beta \) and has units of \( \text{C}^{-1} \). This coefficient is crucial because it helps us predict the volume change of a given material when it experiences a temperature change.

In the exercise, we learned that the coefficient of volume expansion for aluminum is \( 69 \times 10^{-6} \text{C}^{-1} \). This means for every degree Celsius increase, the volume of the aluminum can expand by this factor of its original volume.

When dealing with different materials, knowing their coefficients allows us to determine how much more or less a substance will expand compared to another. The liquid in the can also has a coefficient, which we calculated as \( 8.29 \times 10^{-5} \text{C}^{-1} \). This value being higher than that of aluminum explains why the liquid overflows when heated.
Volume Expansion
Volume expansion occurs when a substance increases in volume as a result of being heated. It's essential to understand that different materials expand at different rates, influenced by their respective coefficients of volume expansion.

For a precise calculation, use the formula: \[\Delta V = \beta V_0 \Delta T\]
  • \( \Delta V \) represents the change in volume.
  • \( \beta \) is the coefficient of volume expansion.
  • \( V_0 \) is the initial volume of the substance.
  • \( \Delta T \) is the change in temperature.
In our scenario, for the aluminum can, the calculated change in volume (\( \Delta V \)) was \( 1.76415 \times 10^{-5} \text{m}^3 \). This change in the can's size partially accommodated the expansion of the liquid.

However, some liquid still spilled over, pointing to a larger volume expansion than that of the container. Thus, when heating a liquid in a container, it's essential to account for both the container's and liquid's expansions.
Temperature Change
Temperature change, denoted as \( \Delta T \), simply describes the difference between the final and initial temperatures. Calculating this is straightforward and necessary for solving many thermal expansion problems.

In the given exercise, the temperature change for the aluminum can and the liquid was:\[\Delta T = 78^{\circ} \text{C} - 5^{\circ} \text{C} = 73^{\circ} \text{C}.\]This value shows how much the temperature increased during heating. It is used in the equation for volume expansion to determine how much a material changes in volume at a certain temperature increase.

Understanding temperature change is pivotal because it directly affects how much a material will expand. A larger temperature increase will typically result in a more significant volume change, which was evident when the liquid expanded dramatically and some spilled from the can.

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Most popular questions from this chapter

An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0{ }^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C} .\) All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

When resting, a person has a metabolic rate of about \(3.0 \times\) \(10^{5}\) joules per hour. The person is submerged neck-deep into a tub containing \(1.2 \times 10^{3} \mathrm{kg}\) of water at \(21.00^{\circ} \mathrm{C} .\) If the heat from the person goes only into the water, find the water temperature after half an hour.

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{kg}\) of blood flows to the body's surface and releases \(2000 \mathrm{J}\) of energy. The blood arriving at the surface has the temperature of the body's interior, \(37.0^{\circ} \mathrm{C} .\) Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

You and your team are given the task of constructing a crude thermometer that covers a temperature range from \(0^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C} .\) You have at your disposal an aluminum rod of length \(L_{0}=2.00 \mathrm{m}\) (at \(\left.T=0^{\circ} \mathrm{C}\right)\) and diameter \(D=0.50 \mathrm{cm},\) and a broken clock that is missing its hour hand, and its longer minute hand is hanging loose and pointing in the six o"clock direction. The long hand of the clock is 8.20 inches long and pivots loosely at the clock's center. You get the idea to mount the rod horizontally so that one end butts against a wall and the other end pushes against the dangling minute hand of the clock. A temperature-induced change in the length of the rod will then be reflected in a change in the angle that the minute hand makes with the vertical (i.e., relative to the six o'clock position). (a) If you want the full temperature range to span the angle between the 6 and 7 markings (uniformly spaced) on the clock, how far from the central pivot should the end of the rod make contact with the minute hand? (b) The current temperature in the room is \(65.0^{\circ} \mathrm{F} .\) At what angle relative to the vertical (the six o'clock position) should the minute hand point at this temperature if it is to point directly at 6 when \(T=0^{\circ} \mathrm{C} ?\) (c) What would be the angular range of your "clock-thermometer" if the rod were made of steel, rather than aluminum? Assume that it is placed at the same position on the minute hand as determined in (a).

The vapor pressure of water at \(10^{\circ} \mathrm{C}\) is \(1300 \mathrm{Pa}\). (a) What percentage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5}\) Pa. (b) What percentage of the total air pressure at \(10^{\circ} \mathrm{C}\) is due to water vapor when the relative humidity is \(100 \% ?\) (c) The vapor pressure of water at \(35^{\circ} \mathrm{C}\) is 5500 Pa. What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} \mathrm{C}\) when the relative humidity was \(100 \% ?\)
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