Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50-min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 m\(^3\) of air in the room. The air has specific heat 1020 J/kg \(\cdot\) K and density 1.20 kg/m\(^3\). If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50-min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 W. What is the temperature rise during 50 min in this case?

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0\(^\circ\)C. How many kilograms of ice at \(-\)20.0\(^\circ\)C must be dropped into the water to make the final temperature of the system 40.0\(^\circ\)C?

Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 C\(^\circ\)? The heat of vaporization of water at body temperature (37\(^\circ\)C) is \(2.42 \times 10{^6} J/kg\). The specific heat of a typical human body is 3480 J/kg \(\cdot\) K (see Exercise 17.25). (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can (355 cm\(^3\)).

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is 0.780 kg, and its temperature increases from 18.55\(^\circ\)C to 22.54\(^\circ\)C. (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free