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A glass flask whose volume is 1000.00 cm\(^3\) at 0.0\(^\circ\)C is completely filled with mercury at this temperature. When flask and mercury are warmed to 55.0\(^\circ\)C, 8.95 cm\(^3\) of mercury overflow. If the coefficient of volume expansion of mercury is \(18.0 \times 10{^-}{^5} K{^-}{^1}\), compute the coefficient of volume expansion of the glass.

Short Answer

Expert verified
The coefficient of volume expansion of the glass is approximately \(1.73 \times 10^{-5} K^{-1}\).

Step by step solution

01

Understand the Provided Information

We have a glass flask with a volume of 1000.00 cm to evaluate the expansion of mercury first.
02

Calculate the Expansion of Mercury

The change in volume due to expansion is given by \( \Delta V = V_0 \beta_{mercury} \Delta T \). Here, \( V_0 = 1000.00 \text{ cm}^3 \) is the initial volume, \( \beta_{mercury} = 18.0 \times 10^{-5} K^{-1} \) is the coefficient of volume expansion of mercury, and \( \Delta T = 55.0 - 0.0 = 55.0 \text{ K} \) is the change in temperature. We need to compute the expanded volume of mercury.
03

Substitute Values to Find Change in Mercury Volume

Substitute the known values into the formula: \[\Delta V = 1000.00 \times 18.0 \times 10^{-5} \times 55.0\] Calculate \( \Delta V \) to find how much the volume increases.
04

Calculate the Numerical Result for Mercury Expansion

Perform the calculations: \[\Delta V = 1000.00 \times 18.0 \times 10^{-5} \times 55.0 = 9.9 \text{ cm}^3\] Thus, mercury's volume would increase by 9.9 cm³ due to heating.
05

Determine Flask's Expansion

Since 8.95 cm³ of mercury overflowed, the volume expansion of the flask itself must account for the difference between the mercury expansion and overflow. Therefore,\[V_{flask\_expansion} = 9.9 - 8.95 = 0.95 \text{ cm}^3\].
06

Calculate the Coefficient of Volume Expansion of the Glass

We use the formula \( \Delta V_{flask} = V_0 \beta_{glass} \Delta T \) where \( \Delta V_{flask} = 0.95 \text{ cm}^3 \), \( V_0 = 1000.00 \text{ cm}^3 \), and \( \Delta T = 55.0 \text{ K} \). Solving for \( \beta_{glass} \):\[\beta_{glass} = \frac{0.95}{1000.00 \times 55.0}\]. Substitute the known values to find \( \beta_{glass} \).
07

Solve the Expression for the Glass Expansion Coefficient

Perform the calculation for \( \beta_{glass} \): \[\beta_{glass} = \frac{0.95}{1000.00 \times 55.0} = \frac{0.95}{55000} \approx 1.73 \times 10^{-5} K^{-1}\].

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Key Concepts

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

Thermal Expansion
Thermal Expansion refers to the tendency of a material to change its volume in response to a change in temperature. When materials are heated, their particles begin to move more vigorously. This increased motion causes the material to expand, which is particularly noticeable in solids, liquids, and gases.
In the case of this exercise, the glass flask and the mercury inside it both experience thermal expansion. Since the coefficient of volume expansion differs between different materials, not all components expand at the same rate when subjected to the same temperature change. This differing expansion is the reason we observe overflow in the mercury from the flask.
  • The coefficient of volume expansion (\(\beta)\) is a property that quantifies how much a material's volume changes per degree change in temperature.
  • In the formula \(\Delta V = V_0 \beta \Delta T\), \\(\Delta V\) is the change in volume, \\(V_0\) is the initial volume, \\(\beta\) is the coefficient of volume expansion, and\(\Delta T\) is the temperature change.
In practical terms, engineers and designers must consider thermal expansion in applications such as building bridges, where temperatures can vary dramatically between seasons.
Physics Problem Solving
In tackling physics problems, particularly those involving thermal expansion like the one in this exercise, the approach involves understanding given information and applying relevant formulas to find unknowns.
The step-by-step process follows a structured approach:
  • Understand the problem: Identify what is known and what needs to be found. Here, we know the initial volume of the flask and mercury, the overflow amount, and the coefficient of mercury's expansion.
  • Calculate known variables: Calculate the change in volume (\(\Delta V)\) for mercury using the formula \\(\Delta V = V_0 \beta \Delta T\). This helps determine how much the volume of mercury increases when heated.
  • Analyze results: By calculating the difference in volume expansions between the mercury and the flask, you find how much the mercury overflows due to expansion.
  • Solve for unknowns: Determine the coefficient of volume expansion of the glass by rearranging formulas accordingly and substituting known quantities.
This logical way of breaking down the problem helps to simplify complex physics concepts, ensuring that all parts are understood and solved correctly.
Temperature Change
Temperature Change is a critical element affecting thermal expansion. In this exercise, it's calculated as the difference between the final and initial temperatures of the system. This change directly influences how much the volume of the materials expand.
Understanding \(\Delta T\) (the change in temperature) is key in applying the thermal expansion formulas as it helps quantify how significant the expansion will be.
  • \(\Delta T\) for our exercise is 55.0 K, calculated simply as \\(55.0 - 0.0\).
  • Temperature changes affect different materials in distinct ways depending on their expansion coefficients.
  • Accurate measurement of temperature change ensures precise assessments of how materials like glass and mercury will react when heated.
In practical scenarios, knowing how materials expand with temperature change is crucial for designing systems that accommodate this expansion, ensuring that they maintain structural integrity and functionality across various temperatures.

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Most popular questions from this chapter

BIO Conduction Through the Skin. The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0\(^\circ\)C and that the outer surface of the skin is at 30.0\(^\circ\)C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m\(^2\) and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person’s skin?

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is -20\(^\circ\)C, what amount of heat is needed to warm to body temperature (37\(^\circ\)C) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020 J / kg \(\cdot\) K and that 1.0 L of air has mass \(1.3 \times 10{^-}{^3} kg\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student’s body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to 44\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0\(^\circ\)F to 45.0\(^\circ\)F in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was 44.0\(^\circ\)F on January 23, 1916. The next day the temperature plummeted to -56\(^\circ\)F. What was the temperature change in Celsius degrees?

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