/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 You are making pesto for your pa... [FREE SOLUTION] | 91Ó°ÊÓ

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You are making pesto for your pasta and have a cylindrical measuring cup \(10.0 \mathrm{~cm}\) high made of ordinary glass \(\left[\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) that is filled with olive oil \(\left[\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) to a height of \(3.00 \mathrm{~mm}\) below the top of the cup. Initially, the cup and oil are at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\). 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 precise calculation would require specific radius for the measuring cup and oil. Once that is provided, apply the above steps to get the desired temperature at which the oil will start spilling.

Step by step solution

01

Calculate the initial volumes

Calculate the initial volume of the oil and the available volume for the oil in the cup. The volume of a cylinder is given by \(V=\pi r^2 h\). For the oil, the height is given as 10 cm minus 3 mm = 9.97 cm. For the cup, the height is the full 10 cm.
02

Apply the volumetric expansion formula for the final volumes

Apply the volumetric expansion formula \(V_{final}=V_{initial}(1+\beta \Delta T)\) separately to the oil and the cup to give us their respective final volumes. The temperature will start from 22 degrees Celsius and increase, so the change in temperature \(\Delta T\) will be the final temperature minus 22. We are asked to find the temperature at which the oil starts to spill, indicating that the final volume of the oil is equal to the final volume of the cup. Hence, we set the two final volume equations equal to each other.
03

Solve for the final temperature

Solve the resulting equation for the final temperature. This will involve some algebra and possibly using the quadratic equation if it simplifies to such. This will yield the temperature in Celsius at which the oil will start 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.

Volumetric Expansion
Volumetric expansion is a concept in physics that describes how the volume of a material changes with temperature. As a substance heats up, its molecules move faster and tend to spread apart, causing it to expand. This phenomenon is particularly important when dealing with liquids and gases, which typically expand more than solids. When calculating volumetric expansion, the following formula is used:
  • \[ V_{\text{final}} = V_{\text{initial}} (1 + \beta \Delta T) \]
Here, \( V_{\text{final}} \) is the final volume, \( V_{\text{initial}} \) is the initial volume, \( \beta \) is the coefficient of volumetric expansion, and \( \Delta T \) is the change in temperature.
In the case of olive oil heating up, the volumetric expansion formula helps us determine at what temperature the volume of oil will equal the maximum volume the glass cup can hold. Understanding this is key to predicting when and how materials will behave under temperature changes.
Coefficient of Thermal Expansion
The coefficient of thermal expansion is a crucial property that tells us how much a material will expand per degree of temperature change. It is typically denoted as \( \beta \) or \( \alpha \) when discussing volumetric or linear expansion, respectively.
Materials have different expansion coefficients due to their unique molecular structures.
  • For example, glass has a smaller coefficient of thermal expansion \((2.7 \times 10^{-5} \, \text{°C}^{-1})\) compared to olive oil \((6.8 \times 10^{-4} \, \text{°C}^{-1})\).
  • This means olive oil will expand more quickly than glass as temperature increases.
Knowing the coefficient of expansion is especially important in applications where materials will be subjected to temperature changes, as it allows us to anticipate and counteract potential issues like spilling or breakage.
Cylindrical Volume Calculation
The calculation of volume for objects, such as the cylindrical measuring cup in this exercise, involves understanding geometric formulas. For a cylinder, the volume is calculated using:
  • \[ V = \pi r^2 h \]
Where \( r \) is the radius of the cylinder's base, and \( h \) is the height.
In the context of the exercise, you first need to calculate the volume of olive oil using its initial conditions, where the height is reduced by 3 mm from the total height of the cup.
The remaining capacity of the measuring cup before expansion also needs to be recalculated to ensure there's no spillage. Ensuring your calculations of cylindrical volumes are precise is key to predicting the outcome when both the cup and its contents are heated.

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Most popular questions from this chapter

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{~K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{~K}) ;\) (c) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{~K}\right)\)

At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye's \(T^{3}\) law: $$ C=k \frac{T^{3}}{\theta^{3}} $$ where \(k=1940 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) and \(\theta=281 \mathrm{~K}\). (a) How much heat is required to raise the temperature of \(1.50 \mathrm{~mol}\) of rock salt from \(10.0 \mathrm{~K}\) to \(40.0 \mathrm{~K} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at \(40.0 \mathrm{~K} ?\)

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped \(10.0 \mathrm{~K}\). What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and \((\mathrm{b}) \mathrm{C}^{\circ} ?\)

You have \(750 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

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