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Consult Interactive LearningWare 12.1 at for help in solving this problem. 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 returns, 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 \( 7.77 \times 10^{-6} \; \text{m}^3 \) of coffee spilled out.

Step by step solution

01

Understanding Volume Expansion

When a liquid is heated, its volume changes according to 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

Identify Given Values

From the problem, we know:- Initial volume \( V_0 = 0.50 \times 10^{-3} \; \text{m}^3 \).- Initial temperature \( T_i = 18^{\circ} \text{C} \).- Final temperature \( T_f = 92^{\circ} \text{C} \).- \( \beta \approx 2.1 \times 10^{-4} \; \text{C}^{-1} \) for water (and coffee as stated).
03

Calculate Temperature Change

The change in temperature \( \Delta T \) is the difference between the final and initial temperatures: \[ \Delta T = T_f - T_i = 92^{\circ} \text{C} - 18^{\circ} \text{C} = 74^{\circ} \text{C} \]
04

Calculate Volume Change

Substitute the known values into the volume expansion formula:\[ \Delta V = \beta V_0 \Delta T \]\[ \Delta V = (2.1 \times 10^{-4} \; \text{C}^{-1})(0.50 \times 10^{-3} \; \text{m}^3)(74^{\circ} \text{C}) \]Calculate \( \Delta V \):\[ \Delta V = 7.77 \times 10^{-6} \; \text{m}^3 \]
05

Conclusion

Thus, the amount of coffee that has spilled out of the beaker due to heating is approximately \( 7.77 \times 10^{-6} \; \text{m}^3 \).

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Key Concepts

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

Volume Expansion
Volume expansion is an important concept in thermal physics. It describes how the volume of a substance changes when it is subjected to temperature variations. In liquids, this phenomenon is particularly noticeable.
When heated, particles in a liquid gain energy and move apart, causing the liquid to occupy more space. This leads to an increase in volume, known as thermal expansion. The formula for volume expansion 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.
This principle explains why the coffee in the beaker spills over when heated.
Coefficient of Volume Expansion
The coefficient of volume expansion (\( \beta \)) is a crucial parameter in understanding how much a substance will expand when its temperature changes. Different materials have different coefficients, which is why they expand at different rates.
For liquids like water and coffee, the coefficient of volume expansion is approximately \( 2.1 \times 10^{-4} \; \text{C}^{-1} \). This value indicates how much the volume of the liquid will increase for each degree Celsius rise in temperature.
Using the coefficient, we can predict the volume increase of the liquid from a temperature change by plugging into:
\[ \Delta V = \beta V_0 \Delta T \]Understanding this coefficient helps in solving practical problems like the spilling coffee exercise, showing how predictive and useful this concept is.
Thermal Physics
Thermal physics explains how temperature and heat energy affect materials. It's the branch of physics that studies the energy transfer between different systems. A fundamental aspect here is understanding how changes in temperature can alter physical properties like volume.
When you heat a material, you essentially supply energy to its particles. They begin to vibrate faster, increasing the space they occupy. This is evident in phenomena like melting, evaporation, and of course, thermal expansion.
In the context of the exercise, thermal physics principles show how heat causes the coffee to expand. Thus, by understanding these concepts, one can predict and explain daily observations like the spilling of liquids upon heating.
Temperature Change
Temperature change is the variation in temperature that an object or substance experiences over time. It's a fundamental factor in various physical changes in materials.
In our coffee example, the temperature rise is crucial for predicting the expansion of the liquid. We calculate this change using the formula:
\[ \Delta T = T_f - T_i \]where:
  • \( T_f \) is the final temperature, and
  • \( T_i \) is the initial temperature.
Knowing \( \Delta T \), you can use it in the volume expansion formula to find out how much the volume will increase.
Therefore, understanding and calculating temperature change is key to solving problems involving thermal expansion.

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Most popular questions from this chapter

When you take a bath, how many kilograms of hot water \(\left(49.0^{\circ} \mathrm{C}\right)\) must you mix with cold water \(\left(13.0^{\circ} \mathrm{C}\right)\) so that the temperature of the bath is \(36.0^{\circ} \mathrm{C} ?\) The total mass of water (hot plus cold) is \(191 \mathrm{~kg}\). Ignore any heat flow between the water and its external surroundings.

Equal masses of two different liquids have the same temperature of \(25.0^{\circ} \mathrm{C}\). Liquid A has a freezing point of \(-68.0{ }^{\circ} \mathrm{C}\) and a specific heat capacity of \(1850 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) Liquid \(\mathrm{B}\) has a freezing point of \(-96.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(2670 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference \(L_{\mathrm{f}, \mathrm{A}}-L_{\mathrm{f}, \mathrm{B}}\) between the latent heats of fusion for these liquids.

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{~kg}\) of blood flows to the surface of the body and releases \(2000 \mathrm{~J}\) of energy. The blood arriving at the surface has the temperature of the body interior, \(37.0^{\circ} \mathrm{C}\). Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

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 piece of glass has a temperature of \(83.0^{\circ} \mathrm{C}\). Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ}\) C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

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