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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

Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is 2.5 gal/min (9.46 L/min) with the tap water being heated from 50\(^\circ\)F (10\(^\circ\)C) to 120\(^\circ\)F (49\(^\circ\)C) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.

On a cool (4.0\(^\circ\)C) Saturday morning, a pilot fills the fuel tanks of her Pitts S-2C (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again 4.0\(^\circ\)C, she checks the fuel level and finds only 103.4 L of gasoline in the aluminum tanks. She realizes that it was hot on Saturday afternoon and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in \(^\circ\)C) of the fuel and the tank on Saturday afternoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10{^-}{^4} K{^-}{^1}\). (b) To have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

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