15.13

Concepts & Calculations

The first law of thermodynamics is basically a restatement of the conservation-of-energy principle in terms of heat and work. Example 14 emphasizes this important fact by showing that the conservation principle and the first law provide the same approach to a problem. In addition, the example reviews the concept of latent heat of sublimation (see Section 12.8) and the ideal gas law (see Section 14.2).

| Concepts & Calculations Example 14 | | | The Sublimation of Zinc

The sublimation of zinc () takes place at a temperature of , and the latent heat of sublimation is . The pressure remains constant during the sublimation. Assume that the zinc vapor can be treated as a monatomic ideal gas and that the volume of solid zinc is negligible compared to the corresponding vapor. What is the change in the internal energy of the zinc when 1.50 kg of zinc sublimates?
Concept Questions and Answers What is sublimation and what is the latent heat of sublimation?

Answer Sublimation is the process whereby a solid phase changes directly into a gas phase in response to the input of heat. The heat per kilogram needed to cause the phase change is called the latent heat of sublimation Ls. The heat Q needed to bring about the sublimation of a mass m of solid material is given by Equation 12.5 as .

When a solid phase changes to a gas phase, does the volume of the material increase or decrease, and by how much?

Answer Gases, in general, have greater volumes than solids do, so the volume of the material increases. The increase in volume is . Since the volume of the solid Vsolid is negligibly small in comparison to the volume of the gas Vgas, we have . Using the ideal gas law as given in Equation 14.1, it follows that , so that . In this result, n is the number of moles of material, R is the universal gas constant, and T is the Kelvin temperature.

As the material changes from a solid to a gas, does it do work on the environment or does the environment do work on it? How much work is involved?

Answer To make room for itself, the expanding material must push against the environment and, in so doing, does work on the environment. Since the pressure remains constant, the work done by the material is given by Equation 15.2 as . Since , the work becomes .

In this problem we begin with heat Q and realize that it is used for two purposes: First, it makes the solid change into a gas, which entails a change in the internal energy of the material, . Second, it allows the expanding material to do work W on the environment. According to the conservation-of-energy principle, how is Q related to and W?

Answer According to the conservation-of-energy principle, energy can neither be created nor destroyed, but can only be converted from one form to another (see Section 6.8). Therefore, part of the heat Q is used for and part for W, with the result that .

According to the first law of thermodynamics, how is Q related to and W?

Answer As indicated in Equation 15.1, the first law of thermodynamics is . Rearranging this equation gives , which is identical to the result obtained from the conservation-of-energy principle.

Solution Using the facts that , we have that

Solving for gives

In this result, n is the number of moles of the ideal gas. According to the discussion in Section 14.1, the number of moles of gaseous zinc is the mass m of the sample divided by the mass per mole of zinc or n = m/(0.0654 kg/mol). Therefore, we find

Heat engines can be used to perform work, as we have seen in this chapter. The concept of work, however, was first introduced in Chapter 6, along with the idea of kinetic energy and the work–energy theorem. The next example reviews some of the main features of heat engines, as well as kinetic energy and the work–energy theorem.

| Concepts & Calculations Example 15 | | | The Work–Energy Theorem

Each of two Carnot engines uses the same cold reservoir at a temperature of 275 K for its exhaust heat. Each engine receives 1450 J of input heat. The work from either of these engines is used to drive a pulley arrangement that uses a rope to accelerate a 125-kg crate from rest along a horizontal frictionless surface, as Figure 15-24 suggests. With engine 1 the crate attains a speed of 2.00 m/s, while with engine 2 it attains a speed of 3.00 m/s. Find the temperature of the hot reservoir for each engine.

Figure 15-24 With the aid of pulleys and a rope, a Carnot engine provides the work that is used to accelerate the crate from rest along a horizontal frictionless surface. See Concepts & Calculations Example 15.

Concept Questions and Answers With which engine is the change in the crate’s kinetic energy greater?

Answer The change is greater with engine 2. Kinetic energy is , according to Equation 6.2, where m is the mass of the crate and v is its speed. The change in the kinetic energy is the final minus the initial value, or KEf − KE0. Since the crate starts from rest, it has zero initial kinetic energy. Thus, the change is equal to the final kinetic energy. Since engine 2 gives the crate the greater final speed, it causes the greater change in kinetic energy.

Which engine does more work?

Answer The work-energy theorem, as stated in Equation 6.3, indicates that the net work done on an object equals the change in the object’s kinetic energy, or . The net work is the work done by the net force. In Figure 15-24 the surface is horizontal, and the crate does not leave it. Therefore, the upward normal force that the surface applies to the crate must balance the downward weight of the crate. Furthermore, the surface is frictionless, so there is no friction force. The net force acting on the crate, then, consists of the single force due to the tension in the rope, which arises from the action of the engine. Thus, the work done by the engine is, in fact, the net work done on the crate. But we know that engine 2 causes the crate’s kinetic energy to change by the greater amount, so that engine must do more work.

For which engine is the temperature of the hot reservoir greater?

Answer The temperature of the hot reservoir for engine 2 is greater. We know that engine 2 does more work, but each engine receives the same 1450 J of input heat. Therefore, engine 2 derives more work from the input heat. In other words, it is more efficient. But the efficiency of a Carnot engine depends only on the Kelvin temperatures of its hot and cold reservoirs. Since both engines use the same cold reservoir whose temperature is 275 K, only the temperatures of the hot reservoirs are different. Higher temperatures for the hot reservoir are associated with greater efficiencies, so the temperature of the hot reservoir for engine 2 is greater.

Solution According to Equation 15.11, the efficiency e of a heat engine is the magnitude of the work |W| divided by the magnitude of the input heat |QH|, or . According to Equation 15.15, the efficiency of a Carnot engine is , where TC and TH are, respectively, the Kelvin temperatures of the cold and hot reservoirs. Combining these two equations, we have

But |W| is the magnitude of the net work done on the crate, and it equals the change in the crate’s kinetic energy, or , according to Equations 6.2 and 6.3. With this substitution, the efficiency expression becomes

Solving for the temperature TH, we find

As expected, the value for TH for engine 2 is greater:

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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