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During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} \mathrm{~m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} \mathrm{C}\), and the beaker is filled to the brim. A short time later when the student retums, the temperature has risen to \(92^{\circ} \mathrm{C}\). The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?

Short Answer

Expert verified
Approximately 7.64 x 10^-6 cubic meters of coffee spills.

Step by step solution

01

Understanding Volume Expansion

The volume expansion of a liquid due to temperature change is given by the formula \( \Delta V = \beta V_0 \Delta T \), 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.
02

Identifying Known Values

The initial volume of the coffee \( V_0 = 0.50 \times 10^{-3} \) m鲁. The initial temperature \( T_i = 18^{\circ}C \) and the final temperature \( T_f = 92^{\circ}C \) gives a temperature change \( \Delta T = T_f - T_i = 74^{\circ}C \). The coefficient of volume expansion for water (and coffee, by assumption) is \( \beta = 207 \times 10^{-6} \) 掳C鈦宦.
03

Calculating Change in Volume

Substitute the known values into the volume expansion equation: \( \Delta V = \beta V_0 \Delta T = 207 \times 10^{-6} \times 0.50 \times 10^{-3} \times 74 \).
04

Performing the Calculation

Calculate the expression: \( \Delta V = 207 \times 10^{-6} \times 0.50 \times 10^{-3} \times 74 = 7.6419 \times 10^{-6} \) m鲁.

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Key Concepts

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

Coefficient of Volume Expansion
In physics, the coefficient of volume expansion is a measure of how much a substance expands when its temperature increases. Materials change size when heated because their molecules move faster and tend to move apart. - The coefficient, denoted as \( \beta \), quantifies this expansion. - It is expressed in units of per degree Celsius (掳C鈦宦). - For liquids like water or coffee, this coefficient is relatively small, reflecting that they do expand but not excessively.In the context of our exercise, the coefficient of volume expansion for coffee (or water) is \( \beta = 207 \times 10^{-6} \) 掳C鈦宦. This means that for every degree Celsius rise in temperature, the volume of coffee will expand proportionally by this coefficient. This property is crucial in calculating how much liquid will spill out when heated in a full container.
Temperature Change
Temperature change is a straightforward concept that refers to the difference between the final and initial temperatures of a substance. It is usually the result of transferring energy to or from the substance, leading to a measurable change in temperature. - The temperature change, represented as \( \Delta T \), is calculated by subtracting the initial temperature from the final temperature.- In our exercise, the initial temperature was \( 18^{\circ} \mathrm{C} \) and the final temperature was \( 92^{\circ} \mathrm{C} \).- Therefore, the temperature change \( \Delta T = 92^{\circ} \mathrm{C} - 18^{\circ} \mathrm{C} = 74^{\circ} \mathrm{C} \).The temperature change informs us about how much the environment around the liquid and beaker has been altered, linking directly to how much the liquid will expand.
Initial Volume
The initial volume of a substance is a fundamental factor in determining its behavior when subjected to temperature variations. - It denotes the amount of space occupied by the substance before any external changes, such as heating or cooling.- In our case, the initial volume of the coffee is stated as \( 0.50 \times 10^{-3} \) m鲁. Knowing the initial volume helps us calculate the potential change in volume when temperature changes. It is a key input for applying the volume expansion formula, necessary to determine how much coffee will spill from the beaker during heating.
Volume Calculation
Volume calculation for thermal expansion involves using a specific mathematical formula to find out how much a liquid's volume changes with temperature.- The formula is: \( \Delta V = \beta V_0 \Delta T \), 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 temperature change.In our specific example:
  • We substitute the values: \( \beta = 207 \times 10^{-6} \) 掳C鈦宦, \( V_0 = 0.50 \times 10^{-3} \) m鲁, and \( \Delta T = 74 \) 掳C.
  • Performing the calculation yields: \( \Delta V = 207 \times 10^{-6} \times 0.50 \times 10^{-3} \times 74 = 7.6419 \times 10^{-6} \) m鲁.
This result shows the additional volume the coffee gains due to the increase in temperature, indicating how much coffee will spill out.

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Most popular questions from this chapter

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

To help keep his barn warm on cold days, a farmer stores \(840 \mathrm{~kg}\) of solar-heated water \(\left(L_{\mathrm{f}}=3.35 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right)\) in barrels. For how many hours would a \(2.0-\mathrm{kW}\) electric space heater have to operate to provide the same amount of heat as the water does when it cools from 10.0 to \(0.0^{\circ} \mathrm{C}\) and completely freezes?

A copper-constantan thermocouple can generate a voltage of \(4.75 \times 10^{-3}\) volts when the temperature of the hot junction is \(110.0{ }^{\circ} \mathrm{C}\) and the reference junction is kept at \(0.0^{\circ}\) C. If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

A steel ruler is calibrated to read true at \(20.0^{\circ} \mathrm{C}\). A draftsman uses the ruler at \(40.0^{\circ} \mathrm{C}\) to draw a line on a \(40.0^{\circ} \mathrm{C}\) copper plate. As indicated on the warm ruler, the length of the line is \(0.50 \mathrm{~m}\). To what temperature should the plate be cooled, such that the length of the line truly becomes \(0.50 \mathrm{~m}\) ?

Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the water becomes hot. The heating system of a house has \(76 \mathrm{~m}\) of copper pipe whose inside radius is \(9.5 \times 10^{-3} \mathrm{~m}\). When the water and pipe are heated from 24 to \(78^{\circ} \mathrm{C}\), what must be the minimum volume of the reservoir tank to hold the overflow of water?

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