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

Short Answer

Expert verified
The coefficient of volume expansion of the glass is approximately \( 1.727 \times 10^{-5} \mathrm{K}^{-1} \).

Step by step solution

01

Understand the Problem

We have a glass flask that is initially filled with mercury at a volume of 1000.00 \( \mathrm{cm}^3 \) at \( 0.0^{\circ} \mathrm{C} \). When both the flask and the mercury are heated to \( 55.0^{\circ} \mathrm{C} \), 8.95 \( \mathrm{cm}^3 \) of mercury overflows. We need to use this information to find the coefficient of volume expansion of the glass flask.
02

Formula of Volume Expansion

The formula for volume expansion due to temperature change is given by:\[\Delta V = \beta V_0 \Delta T\]where \( \Delta V \) is the change in volume, \( \beta \) is the coefficient of volume expansion, \( V_0 \) is the initial volume, and \( \Delta T \) is the change in temperature.
03

Calculate Mercury Expansion

For mercury, use the formula of volume expansion:\[\Delta V_{\text{mercury}} = \beta_{\text{mercury}} V_{0,\text{mercury}} \Delta T\]Given \( \beta_{\text{mercury}} = 18.0 \times 10^{-5} \mathrm{K}^{-1} \), \( V_{0,\text{mercury}} = 1000.00 \, \mathrm{cm}^3 \), and \( \Delta T = 55.0 \degree C - 0.0 \degree C = 55.0 \degree C \), calculate \( \Delta V_{\text{mercury}} \):\[\Delta V_{\text{mercury}} = 18.0 \times 10^{-5} \times 1000.00 \times 55.0 = 9.90 \mathrm{cm}^3\]
04

Determine Expansion of Glass

Since 8.95 \( \mathrm{cm}^3 \) of mercury overflowed, the volume by which the glass flask expanded is:\[\Delta V_{\text{glass}} = \Delta V_{\text{mercury}} - \text{overflowed volume} = 9.90 \mathrm{cm}^3 - 8.95 \mathrm{cm}^3 = 0.95 \mathrm{cm}^3\]
05

Calculate Coefficient of Expansion for Glass

Using the expansion formula for the glass:\[\Delta V_{\text{glass}} = \beta_{\text{glass}} V_{0,\text{glass}} \Delta T\]we have:\[0.95 \mathrm{cm}^3 = \beta_{\text{glass}} \times 1000.00 \mathrm{cm}^3 \times 55.0 \degree C\]Solving for \( \beta_{\text{glass}} \):\[\beta_{\text{glass}} = \frac{0.95}{1000.00 \times 55.0} = \frac{0.95}{55000} = 1.727 \times 10^{-5} \mathrm{K}^{-1}\]
06

Conclusion

Therefore, the coefficient of volume expansion for the glass flask is found to be approximately \( 1.727 \times 10^{-5} \mathrm{K}^{-1} \).

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Key Concepts

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

Thermal Expansion
Thermal expansion is a fundamental concept in physics that describes how the volume of a material changes with temperature. When substances are heated, their particles move faster and tend to occupy more space, resulting in an increase in volume. This is because as temperature rises, the kinetic energy of the particles also increases, causing them to vibrate more vigorously and push against each other with greater force.

Materials have specific thermal expansion coefficients, which quantify the change in volume per degree of temperature change. The formula for volume expansion is given as:\[\Delta V = \beta V_0 \Delta T\]where:
  • \( \Delta V \) is 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.
This coefficient is particularly important when designing systems and structures that will be exposed to temperature variations, as it helps in predicting how much the material will expand or contract.

Understanding thermal expansion is essential when solving problems related to volume change due to temperature shifts, as it provides a clear framework to quantify and predict the behavior of materials under thermal stress.
Volume Change
Volume change refers to the alteration in the size of an object resulting from temperature fluctuations. In the exercise provided, the glass flask and the mercury both experience a rise in temperature, leading to their volume changing. However, because they expand at different rates, some mercury overflows from the flask.

It's crucial to recognize the concept of relative expansion when dealing with composite systems like a container and its contents. If a container and its contents (e.g., a liquid) have different coefficients of volume expansion, their relative expansion needs to be considered to calculate overflow or space left in the container.
In our example, the amount of mercury overflow directly relates to the discrepancy in expansion between the mercury and the glass flask. Calculating the overflow involves understanding the distinct expansion behaviors:
  • The initial expansion of mercury calculated using its specific volume expansion coefficient.
  • Determining the expansion of the glass that directly affects how much space remains for the mercury.
Thus, the difference in their expansion helps in identifying the overflowed volume, and in turn lets us find the unknown expansion coefficient of the glass.
Physics Problem Solving
Physics problem solving often requires a structured approach to simplify complex problems, like the one we are addressing involving thermal expansion and volume change. Here's how you tackle these problems sequentially:

1. **Understanding the Problem:** Carefully read the problem to gather all necessary information and identify what's being asked. Know the initial conditions, the changes applied, and the final observations like in our example with the flask and mercury.
2. **Using Relevant Formulas:** Apply the correct equations to connect given data with the unknowns. Here, the volume expansion formula plays a central role. Understanding each term in the formula helps in applying it correctly to solve for unknown quantities.
3. **Carrying Out Calculations:** Calculate the necessary interim results, such as the expanded volume of mercury, before solving for the final unknown like the coefficient of expansion for the glass.
4. **Interpreting Results:** Analyze the outcome to ensure it makes logical sense in terms of the physical context. Ensure dimensions are consistent, and review each step to avoid calculation errors.

Formulating these steps methodically can greatly aid in solving physics problems accurately and efficiently, turning a complex thermal expansion scenario into a comprehensible solution.

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Most popular questions from this chapter

An asteroid with a diameter of 10 \(\mathrm{km}\) and a mass of \(2.60 \times 10^{15} \mathrm{kg}\) impacts the earth at a speed of \(32.0 \mathrm{km} / \mathrm{s},\) landing in the Pacific Ocean. If 1.00\(\%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(10.0^{\circ} \mathrm{C}\) ), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{kg} .\) )

A Walk in the Sun. Consider a poor lost soul walking at 5 \(\mathrm{km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of \(280 \mathrm{W},\) and almost all of this energy is con- verted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\text { skin }}\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is \(54 \mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text { skin }}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\mathrm{air}}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text { skin }}\) is \(36^{\circ} \mathrm{C} ;\) (iii) the skin absorbs radiant energy from the sun at a rate of 1400 \(\mathrm{W} / \mathrm{m}^{2}\) ; (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C}\) . (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C}\) . Which mechanism is the most important? (b) At what rate (in \(\mathrm{L} / \mathrm{h} )\) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\)) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 \(\mathrm{m}^{2} .\) What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C}\) . By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C}\) ?

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove. The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2} .\) The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

Jogging in the Heat of the Day. You have probably seen people jogging in extremely hot weather and wondered Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 \(\mathrm{kg}\) and surface area 1.85 \(\mathrm{m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{W}, 80 \%\) of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\) . (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0 \(^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\) ? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\). must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}.\)

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