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An engine with an efficiency of 0.30 absorbs \(500 \mathrm{J}\) of heat per cycle. (a) How much work does it perform per cycle? (b) How much heat does it discharge per cycle?

Short Answer

Expert verified
(a) The engine performs \(150 J\) of work per cycle. (b) The engine discharges \(350 J\) of heat per cycle.

Step by step solution

01

Understand the efficiency formula

The efficiency (E) of an engine is given by the following formula: \(E = \frac{W}{Q_h}\) Where: \(E\) = efficiency \(W\) = work done by the engine per cycle \(Q_h\) = heat absorbed per cycle
02

Calculate the work done per cycle

We are given \(E = 0.30\) and \(Q_h = 500 J\). Using the efficiency formula, we can solve for work done per cycle (W): \(0.30 = \frac{W}{500 J}\) Now, we will isolate W: \(W = 0.30 \times 500 J\) \(W = 150 J\) So, the engine performs \(150 J\) of work per cycle.
03

Understand the heat formula

The heat discharged (\(Q_c\)) by the engine can be calculated by knowing the heat absorbed and the work done per cycle: \(Q_c = Q_h - W\) Where: \(Q_h\) = heat absorbed per cycle \(W\) = work done by the engine per cycle \(Q_c\) = heat discharged per cycle
04

Calculate heat discharged per cycle

We have calculated the work done per cycle in Step 2, and we know the heat absorbed per cycle. Plug these values into the heat formula: \(Q_c = 500 J - 150 J\) \(Q_c = 350 J\) So, the engine discharges \(350 J\) of heat per cycle. In conclusion, the engine performs (a) 150 J of work per cycle and (b) discharges 350 J of heat per cycle.

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Key Concepts

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

Heat Engine
Just like the heart pumps blood to power our bodies, heat engines are devices that convert heat energy into mechanical work, pumping life into our machines. They are essential in various applications, powering everything from vehicles to electric power plants.

Think of a heat engine as a magical box where you pour heat at one end (the high-temperature source), and it churns out work, like spinning a wheel, while also releasing some heat at the other end (the low-temperature sink). The engine we just described in our exercise has an efficiency of 0.30, meaning it converts 30% of the absorbed heat into useful work. The remaining heat is not lost but is simply transformed, part of what is known as the heat engine cycle.

It's crucial to differentiate between high-quality energy, like the work a heat engine produces, and low-quality energy, like the discharged heat, which is less useful for doing work. This distinction is fundamental in understanding not just heat engines, but also the principles of thermodynamics as a whole.
First Law of Thermodynamics
Imagine having a robust piggy bank that can't be broken open to steal the coins. This is similar to the first law of thermodynamics, which, put simply, is about the conservation of energy—akin to a universal piggy bank of energy. It states that energy cannot be created or destroyed, only converted from one form to another or transferred between systems.

In the context of a heat engine, like the one in our exercise, the first law becomes strikingly evident. The engine absorbed 500 J of heat energy per cycle, and this energy must go somewhere. That 'somewhere' is work, amounting to 150 J, and the remaining heat discharged, 350 J. Each joule of energy is accounted for, and thus, the engine obeys the sacred rule of energy conservation. By comprehending this timeless law, students get a holistic view of how energy transformations govern our universe's operations.
Work-Energy Principle
When we talk about the work-energy principle, we're looking at the direct relationship between work and the kinetic energy of an object. In simpler terms, when work is done on an object, there's a corresponding change in its kinetic energy.

The heat engine in the exercise does work by using the absorbed heat energy. According to the work-energy principle, the work done by the engine, 150 J per cycle, should directly relate to the kinetic energy imparted to whatever the engine is moving—an evident practical application of this principle. By understanding the work-energy principle, students can predict and calculate the movement of objects influenced by various forces, essentially decoding a part of the narrative of the physical world.
Conservation of Energy
The notion of the conservation of energy is a grand concept, acting as the backbone for understanding physical processes in our universe. This principle states that the total energy in an isolated system remains constant—it is said to be conserved over time.

In our exercise, the conservation of energy manifests when the energy absorbed by the heat engine doesn't just vanish into thin air. Instead, it's reassigned to different forms—the work that's done, and the leftover heat. The energy output (work plus discharged heat) equals the input (absorbed heat), illustrating the conservation of energy. This conservation is a reassuring truth that provides a consistent framework for analyzing physical situations, ensuring students that even in the advent of complex systems and interactions, some things, like total energy, remain unchanging.

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Most popular questions from this chapter

(a) A 5.0-kg rock at a temperature of \(20^{\circ} \mathrm{C}\) is dropped into a shallow lake also at \(20^{\circ} \mathrm{C}\) from a height of \(1.0 \times 10^{3} \mathrm{m} .\) What is the resulting change in entropy of the universe? (b) If the temperature of the rock is \(100^{\circ} \mathrm{C}\) when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that \(c=860 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) is the specific heat of the rock.

A thermal engine produces 4 MJ of electrical energy while operating between two thermal baths of different temperatures. The working substance of the engine discharges 5 MJ of heat to the cold temperature bath. What is the efficiency of the engine?

The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure \(p_{0}\) as the piston expands, increasing the volume of the cylinder from zero to \(V_{A}\) ii. Adiabatic compression stroke \((A B) .\) The temperature of the mixture rises as the piston compresses it adiabatically from a volume \(V_{\mathrm{A}}\) to \(V_{\mathrm{B}}\) iii. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat \(Q_{1}\) causes the pressure to increase from \(p_{B}\) to \(p_{C}\) at the constant volume \(V_{\mathrm{B}}\left(=V_{\mathrm{C}}\right)\) iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from \(V_{C}\) to \(V_{D}\) This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust \((D A)\). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at \(V_{A}\left(=V_{D}\right)\) Most of the available energy is lost here, as represented by the heat exhaust \(Q_{2}\) vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from \(V_{A}\) to zero drives out the remaining combustion products. (a) Using (i) \(e=W / Q_{1} ;\) (ii) \(W=Q_{1}-Q_{2} ;\) and (iii) \(Q_{1}=n C_{\nu}\left(T_{C}-T_{B}\right), Q_{2}=n C_{\nu}\left(T_{D}-T_{A}\right),\) show that \(e=1-\frac{T_{D}-T_{A}}{T_{C}-T_{B}}\) (b) Use the fact that steps (ii) and (iv) are adiabatic to show that \(e=1-\frac{1}{r^{\gamma-1}}\) where \(r=V_{A} / V_{B}\) The quantity \(r\) is called the compression ratio of the engine. (c) In practice, \(r\) is kept less than around 7 . For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for \(r=6\) and \(\gamma=1.4\) (the value for air), \(e=0.51,\) or an efficiency of \(51 \%\) Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about \(25 \%\) to \(30 \%\)

Sketch a Camot cycle on a temperature-volume diagram.

An engineer must design a refrigerator that does 300 J of work per cycle to extract \(2100 \mathrm{J}\) of heat per cycle from a freezer whose temperature is \(-10^{\circ} \mathrm{C}\). What is the maximum air temperature for which this condition can be met? Is this a reasonable condition to impose on the design?

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