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Chapter 11: Problem 18

In a showdown on the streets of Laredo, the good guy drops a \(5.00-\mathrm{g}\) silver bullet at a temperature of \(20.0^{\circ} \mathrm{C}\) into a \(100-\mathrm{cm}^{3}\) cup of water at \(90.0^{\circ} \mathrm{C}\). Simultaneously, the bad guy drops a \(5.00-\mathrm{g}\) copper bullet at the same initial temperature into an identical cup of water. Which one ends the showdown with the coolest cup of water in the West? Neglect any energy transfer into or away from the container.

Short Answer

Expert verified
The bullet material that results to the lesser final temperature makes the water cooler so that determines who ends the showdown with the coolest cup of water in the West.

Step by step solution

01

Identify and input the given values

First, label and put in the given values: mass of bullet and water (\(m_{bullet}=5.00g\) and \(m_{water}=100cm^{3}\)), initial temp of bullet (\(T_{bullet}=20.0^{\circ}C\)), initial temp of water (\(T_{water}=90.0^{\circ}C\)), specific heats (\(c_{silver}=0.235 J/g^{\circ}C\), \(c_{copper}=0.385 J/g^{\circ}C\), and \(c_{water}=4.186 J/g^{\circ}C\)). Also remember that 1cm^3 of water is equal to 1g, thus \(m_{water}\) equals 100g.
02

Calculate the final temperature for silver bullet

Use the equation to compute for the final temperature (T_f) for silver: \(5.00g*0.235 J/g^{\circ}C*(T_{fSilver}-20.0^{\circ}C) = -100g*4.186 J/g^{\circ}C*(T_{fSilver}-90.0^{\circ}C)\). Solve for \(T_{fSilver}\), and that will be the final temperature of the water if a silver bullet is dropped.
03

Calculate the final temperature for the copper bullet

Use the equation to compute for the final temperature (T_f) for copper: \(5.00g*0.385 J/g^{\circ}C*(T_{fCopper}-20.0^{\circ}C) = -100g*4.186 J/g^{\circ}C*(T_{fCopper}-90.0^{\circ}C)\). Solve for \(T_{fCopper}\), and that will be the final temperature of the water if a copper bullet is dropped.
04

Compare the final temperatures

Compare the calculated final temperatures from Step 2 and Step 3. The lesser final temperature will indicate which bullet made the water cooler, and that will answer the exercise.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property of a material that helps us understand how much heat energy is required to change the temperature of a certain mass of the substance by 1 degree Celsius. It is often denoted by the symbol \(c\) and measured in units of \(J/g^{\circ}C\).

Every substance has a unique specific heat capacity. In our exercise, we are comparing silver, copper, and water.

  • Silver has a specific heat capacity of \(0.235 J/g^{\circ}C\).
  • Copper measures \(0.385 J/g^{\circ}C\).
  • Water, however, has a much higher specific heat capacity of \(4.186 J/g^{\circ}C\).
This means that water requires more energy to see the same temperature change compared to silver and copper. Understanding specific heat helps determine how quickly each material will change in temperature when heat is added or removed.
Thermal Equilibrium
Thermal equilibrium is a state reached when two substances involved in heat transfer no longer exchange energy, meaning they have reached the same temperature. This concept is crucial in determining the final temperature in calorimetry problems.

In our exercise, a silver bullet, a copper bullet, and water are initially at different temperatures. As heat flows between the substance with higher initial temperature (the water) and the substances with lower initial temperature (bullets), they aim to reach a final temperature where no net heat is exchanged.

The progress of reaching thermal equilibrium can be modeled through heat transfer calculations, which help us understand how energy flows until equilibrium is achieved.
Temperature Change
Temperature change describes how the heat transfer during a physical process can make a system鈥檚 initial temperature rise or fall. This is important to calculate in the given exercise, where bullets are dropping into water causing changes in temperature.

The key formula for understanding temperature change is \[q = mc\Delta T\] where \(q\) is the heat absorbed or released, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

In the exercise, we calculate how the heat lost by the water compares with the heat gained by the bullets, ultimately allowing us to solve for the unknown: the final temperature of the system.
Calorimetry
Calorimetry is a process that allows us to measure the amount of heat transferred in a chemical reaction or physical change, often using the principles of thermal equilibrium and temperature change.

In this exercise, calorimetry helps us determine the final temperature of water after silver and copper bullets at different initial temperatures are immersed in it.

Steps involved in basic calorimetry calculations include:
  • Identifying mass, specific heat capacity, and initial temperatures of all substances involved.
  • Applying the principle that the heat lost by the water is equal to the heat gained by the bullets.
  • Setting up an equation to solve for the final equilibrium temperature.
By analyzing the calorimetry calculations, students learn not only which bullet results in cooler water but also gain an understanding of heat transfer applications in real systems.

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Most popular questions from this chapter

A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique is to measure the temperature difference between the input and output points of a flowing stream of the liquid while adding energy at a known rate. (a) Start with the equations \(Q=m c(\Delta T)\) and \(m=\rho V\), and show that the rate at which energy is added to the liquid is given by the expression \(\Delta Q / \Delta t=\rho c(\Delta T)(\Delta V / \Delta t)\). (b) In a particular experiment, a liquid of density \(0.72 \mathrm{~g} / \mathrm{cm}^{3}\) flows through the calorimeter at the rate of \(3.5 \mathrm{~cm}^{3} / \mathrm{s}\). At steady state, a temperature difference of \(5.8^{\circ} \mathrm{C}\) is established between the input and output points when energy is supplied at the rate of \(40 \mathrm{~J} / \mathrm{s}\). What is the specific heat of the liquid?

A Styrofoam box has a surface area of \(0.80 \mathrm{~m}^{2}\) and a wall thickness of \(2.0 \mathrm{~cm}\). The temperature of the inner surface is \(5.0^{\circ} \mathrm{C}\), and the outside temperature is \(25^{\circ} \mathrm{C}\). If it takes \(8.0 \mathrm{~h}\) for \(5.0 \mathrm{~kg}\) of ice to melt in the container, determine the thermal conductivity of the Styrofoam.

The apparatus shown in Figure P11.10 was used by Joule to measure the mechanical equivalent of heat. Work is done on the water by a rotating paddle wheel, which is driven by two blocks falling at a constant speed. The temperature of the stirred water increases due to the friction between the water and the paddles. If the energy lost in the bearings and through the walls is neglected, then the loss in potential energy associated with the blocks equals the work done by the paddle wheel on the water. If each block has a mass of \(1.50 \mathrm{~kg}\) and the insulated tank is filled with \(200 \mathrm{~g}\) of water, what is the increase in temperature of the water after the blocks fall through a distance of \(3.00 \mathrm{~m}\) ?

Liquid nitrogen has a boiling point of \(77 \mathrm{~K}\) and a latent heat of vaporization of \(2.01 \times 10^{5} \mathrm{~J} / \mathrm{kg} . \mathrm{A} 25-\mathrm{W}\) electric heating element is immersed in an insulated vessel containing \(25 \mathrm{~L}\) of liquid nitrogen at its boiling point. (a) Describe the energy transformations that occur as power is supplied to the heating element. (b) How many kilograms of nitrogen are boiled away in a period of \(4.0\) hours?

In the summer of 1958 in St. Petersburg, Florida, a new sidewalk was poured near the childhood home of one of the authors. No expansion joints were supplied, and by mid-July the sidewalk had been completely destroyed by thermal expansion and had to be replaced, this time with the important addition of expansion joints! This event is modeled here. A slab of concrete \(4.00 \mathrm{~cm}\) thick, \(1.00 \mathrm{~m}\) long, and \(1.00 \mathrm{~m}\) wide is poured for a sidewalk at an ambient temperature of \(25.0^{\circ} \mathrm{C}\) and allowed to set. The slab is exposed to direct sunlight and placed in a series of such slabs without proper expansion joints, so linear expansion is prevented. (a) Using the linear expansion equation (Eq. 10.4), eliminate \(\Delta L\) from the equation for compressive stress and strain (Eq. 9.3). (b) Use the expression found in part (a) to eliminate \(\Delta T\) from Equation \(11.3\), obtaining a symbolic equation for thermal energy transfer \(Q\). (c) Compute the mass of the concrete slab given that its density is \(2.40 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (d) Concrete has an ultimate compressive strength of \(2.00 \times 10^{7} \mathrm{~Pa}\), specific heat of \(880 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and Young's modulus of \(2.1 \times 10^{10} \mathrm{~Pa}\). How much thermal energy must be transferred to the slab to reach this compressive stress? (e) What temperature change is required? (f) If the Sun delivers \(1.00 \times 10^{3} \mathrm{~W}\) of power to the top surface of the slab and if half the energy, on the average, is absorbed and retained, how long does it take the slab to reach the point at which it is in danger of cracking due to compressive stress?
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