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Chapter 14: Problem 73

\(\bullet\) Pasta time! You are making pesto for your pasta and have a cylindrical measuring cup 10.0 \(\mathrm{cm}\) high made of ordinary glass \(\left(\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) and that is filled with olive oil \(\left(\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) to a height of 1.00 \(\mathrm{mm}\) below the top of the cup. Initially, the cup and oil are at a kitchen temperature of \(22.0^{\circ} \mathrm{C}\) You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

Short Answer

Expert verified
The olive oil will spill at approximately 36.93 掳C.

Step by step solution

01

Determine Key Variables

First, let's identify and define the variables given in the problem: - Height of the measuring cup, \( h_c = 10.0 \ \text{cm} \).- Initial height of the oil inside the cup, \( h_{oil} \), which is \( 1.00 \ \text{mm} \) below the top of the cup, so \( h_{oil} = (10.0 \ \text{cm} - 0.1 \ \text{cm}) \).- Coefficient of volume expansion for the glass, \( \beta_{glass} = 2.7 \times 10^{-5} \ (\text{掳C}^{-1}) \).- Coefficient of volume expansion for the olive oil, \( \beta_{oil} = 6.8 \times 10^{-4} \ (\text{掳C}^{-1}) \).- Initial temperature, \( T_i = 22.0 \ 掳C \).
02

Calculate Volume Changes

The volume expansion can be given by the formula \( \Delta V = \beta V_0 \Delta T \), where \( \Delta T \) is the change in temperature, \( V_0 \) is the initial volume.For the oil, the volume change is \( \Delta V_{oil} = \beta_{oil} V_{0_{oil}} \Delta T \). For the glass/cup, it is \( \Delta V_{glass} = \beta_{glass} V_{0_{cup}} \Delta T \).Since we are dealing with linear dimensions, consider the change in height instead: \( \Delta h_{oil} = \beta_{oil} h_{oil} \Delta T \) and \( \Delta h_{cup} = \beta_{glass} h_c \Delta T \).
03

Equate Volume Expansions

The oil will start spilling out when its expanded volume equals the expanded volume of the cup:\[ h_{oil} + \Delta h_{oil} = h_c + \Delta h_{cup} \]Substitute for \( \Delta h_{oil} \) and \( \Delta h_{cup} \) to find \( \Delta T \):\[ h_{oil} + \beta_{oil} h_{oil} \Delta T = h_c + \beta_{glass} h_c \Delta T \]
04

Solve for Temperature Change

Rearrange the equation to isolate \( \Delta T \):\[ \beta_{oil} h_{oil} \Delta T - \beta_{glass} h_c \Delta T = h_c - h_{oil} \]So,\[ \Delta T = \frac{h_c - h_{oil}}{\beta_{oil} h_{oil} - \beta_{glass} h_c} \]
05

Compute Final Temperature

Substitute the values:\[ \Delta T = \frac{10.0 \ \text{cm} - 9.9 \ \text{cm}}{(6.8 \times 10^{-4})(9.9) - (2.7 \times 10^{-5})(10.0)} \]Calculate \( \Delta T \), then the final temperature: \( T_f = T_i + \Delta T \).
06

Calculation and Conclusion

After computing \( \Delta T \), you find that:\[ \Delta T \approx 14.93 \ 掳C \]Thus, the final temperature is:\[ T_f = 22.0 \ 掳C + 14.93 \ 掳C = 36.93 \ 掳C \]This is the temperature at which the olive oil will begin to spill out of the cup.

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Key Concepts

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

Coefficient of Volume Expansion
The coefficient of volume expansion is a crucial factor in understanding how materials behave when they experience a change in temperature. This coefficient, denoted by \(\beta\), describes how much a material's volume changes with a unit change in temperature. It is typically expressed in units of \((\mathrm{C}^{\circ})^{-1}\).

When we heat materials, they tend to expand. The extent of this expansion depends on the coefficient of volume expansion. For example, in our exercise, the coefficient of volume expansion for olive oil is \(6.8 \times 10^{-4}\ (\mathrm{C}^{\circ})^{-1}\) while for glass, it is \(2.7 \times 10^{-5}\ (\mathrm{C}^{\circ})^{-1}\). The significantly higher coefficient for olive oil compared to glass indicates that olive oil expands much more with temperature changes compared to glass.
  • This concept helps us predict the volume changes in materials when subjected to heating or cooling.
  • The coefficient tells us how susceptible a material is to thermal expansion.
Temperature Change
Temperature change plays a pivotal role in the phenomenon of thermal expansion, as it directly influences the alteration in volume of substances. In the exercise, the initial temperature of the kitchen is given as \(22.0 \, ^{\circ}\mathrm{C}\). You are tasked with determining at what temperature the olive oil will begin to spill over.

The relationship between temperature change and volume change is illustrated by the formula:\[ \Delta V = \beta V_0 \Delta T \]This equation shows that for a given coefficient \(\beta\) and initial volume \(V_0\), the change in volume \(\Delta V\) is proportional to the change in temperature \(\Delta T\).
  • A positive temperature change increases the volume of the substance.
  • The actual temperature at which spillage occurs is calculated by first determining the temperature change (\(\Delta T\)) required for the volume of the oil to exceed that of the cup upon expansion.
Measuring Cup
The measuring cup in this scenario is made of glass and is crucial to understanding the problem's context. With a height of 10 cm, it has been initially filled with olive oil to within 1 mm of the top. It's important to consider the properties of the cup itself, such as its expansion characteristics.

As the temperature of both the cup and the olive oil increases, each will expand. Since both materials have different coefficients of volume expansion, their expansions occur at different rates, and we must consider both to predict any overflow.
  • The initial condition of the cup and oil is important to determine how much thermal expansion can occur before spilling.
  • The cup's expansion (though less than the oil's) still impacts the overall volume available for the oil as the temperature rises.

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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 \(2 \times 10^{15} \mathrm{kg.} )\)

\(\bullet\) Much of the energy of falling water in a waterfall is converted into heat. If all the mechanical energy is converted into heat that stays in the water, how much of a rise in temperature occurs in a 100 m waterfall?

A copper calorimeter can with mass 0.100 \(\mathrm{kg}\) contains 0.160 \(\mathrm{kg}\) of water and 0.018 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?

\(\bullet\) On-demand water heaters. 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 drawback is that energy is wasted because the tank loses heat when it is not in use, and 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 \(/ \min (9.46 \mathrm{L} / \mathrm{min}\) ) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) 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?

\(\bullet\) In a physics lab experiment, a student immersed 200 one-cent coins (each having a mass of 3.00 g) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.240 \(\mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. What was the final temperature of the coins? [One-cent coins are made of a metal alloy-mostly zinc-with a specific heat capacity of 390 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) . ]\)
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