/*! 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 55 A parcel of air over a campfire ... [FREE SOLUTION] | 91Ó°ÊÓ

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A parcel of air over a campfire feels an upward buoyant force because the heated air is less dense than the surrounding air. By estimating the acceleration of the air immediately above a fire, one can estimate the fire's temperature. The mass of a volume \(V\) of air is \(n M_{\text {air }},\) where \(n\) is the number of moles of air molecules in the volume and \(M_{\text {air }}\) is the molar mass of air. The net upward force on a parcel of air above a fire is roughly given by \(\left(m_{\text {out }}-m_{\text {in }}\right) g,\) where \(m_{\text {out }}\) is the mass of a volume of ambient air and \(m_{\text {in }}\) is the mass of a similar volume of air in the hot zone. (a) Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration \(a\) of an air parcel as a function of \(\left(T_{\text {out }} / T_{\text {in }}\right),\) where \(T_{\text {in }}\) is the absolute temperature of the air above the fire and \(T_{\text {out }}\) is the absolute temperature of the ambient air. (b) Rearrange your formula from part (a) to obtain an expression for \(T_{\text {in }}\) as a function of \(T_{\text {out }}\) and \(a\). (c) Based on your experience with campfires, estimate the acceleration of the air above the fire by comparing in your mind the upward trajectory of sparks with the acceleration of falling objects. Thus you can estimate \(a\) as a multiple of \(g .\) (d) Assuming an ambient temperature of \(15^{\circ} \mathrm{C}\), use your formula and your estimate of \(a\) to estimate the temperature of the fire.

Short Answer

Expert verified
We have derived an equation that gives the temperature of the fire based on the observed speed of sparks and the ambient temperature. However, our assumed acceleration value has led to an unphysical temperature, indicating that this assumption was too high. To get a reasonable estimate, the acceleration value assumption needs to be closer to the gravity acceleration.

Step by step solution

01

Derive acceleration \(a\)

First, represent the forces acting on the air parcel above the fire. The net upward force is \(\left(m_{\text {out }}-m_{\text {in }}\right) g\), which by Newton’s second law is also equal to the mass times acceleration: \(m_{\text {in }}a\). The mass of the air volume can be expressed as \(n M_{\text {air}}\) in both cases, where \(n\) is the number of moles in volume \(V\). Using the ideal gas law, \(PV = nRT\), the number of moles \(n\) can be expressed as \(P V / (R T)\). Thus, we rewrite the net force as: \[(P V / (R T_{\text {out}}) M_{\text {air}}) - (P V / (R T_{\text {in}}) M_{\text {air}})] g = (P V / (R T_{\text {in}}) M_{\text {air}}) a\] Solving for \(a\), we get: \(a = g \left[1 - (T_{\text {in}} / T_{\text {out}})\right]\)
02

Find \(T_{\text {in }}\)

Next, rearrange the equation derived in Step 1 to express \(T_{\text {in}}\) as a function of \(T_{\text {out}}\) and \(a\). After rearranging, we have: \(T_{\text {in}} = T_{\text {out}} (1 - a/g)\)
03

Estimate acceleration \(a\)

Estimating the acceleration \(a\) may vary between different people. However, let's assume the acceleration of the sparks is approximately ten times the acceleration due to gravity since they rise quite rapidly, so \(a = 10g\).
04

Estimate fire's temperature

Now, we can calculate \(T_{\text {in }}\) by using the given ambient temperature \(T_{\text {out }} = 15^{\circ} C = 288 K\) and the estimated value of \(a = 10g\). Substituting these into our formula from Step 2, we get: \(T_{\text {in}} = T_{\text {out}} (1 - a/g) = 288(1 - 10) = -2592 K\). However, this is unphysical as absolute temperature cannot be less than zero. This indicates our assumption about the speed of the sparks being ten times the speed of gravity was too high.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in physics and chemistry, useful for relating the pressure, volume, and temperature of a gas with the number of moles. It is formulated as: \(PV = nRT\). Here:
  • \(P\) stands for pressure, often measured in atmospheres or Pascals.
  • \(V\) is the volume, typically in liters.
  • \(n\) is the number of moles, which measures the quantity of gas.
  • \(R\) is the ideal gas constant, valued at approximately \(8.314\) J/(mol K).
  • \(T\) is the absolute temperature, measured in Kelvin.
To understand how the warmth affects the acceleration of air, apply this law. When air is heated, the volume tends to increase if pressure stays constant, resulting in moles occupying more space as temperature rises. This expanded air above a fire feels buoyant compared to colder, denser air. Hence, for equal volumes, the number of moles differs due to the inverse proportionality with temperature in the gas law, thus step into how buoyant forces arise.
Temperature Estimation
Estimating temperature is crucial to determine the behavior of heated air, especially in phenomena like convection above a fire. When you think about the upward acceleration a parcel of air experiences, it indicates how warm the air must be. Using the relation from the previous concept, one can derive the air parcel's acceleration as:
  • The derived equation: \(a = g \left[1 - (T_{\text{in}} / T_{\text{out}})\right]\), provides an understanding of how temperature affects air movement.
  • Solving for \(T_{\text{in}}\) gives us the formula: \(T_{\text{in}} = T_{\text{out}}(1 - a/g)\).
In practical exercises, this formula helps gauge the fire's temperature by knowing the surrounding air temperature and the estimated acceleration. It highlights how much warmer the air above the fire is compared to the ambient temperature, indicating energy transfer from the fire to the air.
Acceleration of Air
Acceleration of air parcels above a heat source, like a campfire, is a captivating observation tied to the buoyancy concept. It represents how fast or slow air rises depending on its temperature relative to surrounding air. This acceleration is influenced by:
  • The buoyant force acting upwards, which is driven by the difference in density between hot and cold air.
  • For practical estimations in exercises, one might consider how sparks rise above a fire compared with known gravitational acceleration \(g\).
Common thought exercises might assume that as sparks fly swiftly, the acceleration could be estimated as a multiple of \(g\), like \(a = 10g\). However, caution must be observed since overestimation can lead to unrealistic results, indicating the necessity to reassess assumptions on speed relative to gravitational effects. Hence, precise temperature and acceleration metrics are essential for sensible realistic modeling.

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Most popular questions from this chapter

The surface of the sun has a temperature of about \(5800 \mathrm{~K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27} \mathrm{~kg} .\) ) (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2},\) where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant (see Example 13.5 of Section 13.3 ). Use Appendix F to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

Helium gas with a volume of \(3.20 \mathrm{~L}\), under a pressure of 0.180 atm and at \(41.0^{\circ} \mathrm{C}\), is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\).

What is one reason the noble gases are preferable to air (which is mostly nitrogen and oxygen) as an insulating material? (a) Noble gases are monatomic, so no rotational modes contribute to their molar heat capacity; (b) noble gases are monatomic, so they have lower molecular masses than do nitrogen and oxygen; (c) molecular radii in noble gases are much larger than those of gases that consist of diatomic molecules; (d) because noble gases are monatomic, they have many more degrees of freedom than do diatomic molecules, and their molar heat capacity is reduced by the number of degrees of freedom.

(a) What is the total translational kinetic energy of the air in an empty room that has dimensions \(8.00 \mathrm{~m} \times 12.00 \mathrm{~m} \times 4.00 \mathrm{~m}\) if the air is treated as an ideal gas at 1.00 atm? (b) What is the speed of a \(2000 \mathrm{~kg}\) automobile if its kinetic energy equals the translational kinetic energy calculated in part (a)?

A person at rest inhales \(0.50 \mathrm{~L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\). The inhaled air is \(21.0 \%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2000 \mathrm{~m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

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