/*! 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 59 Suppose that the gasoline in a c... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that the gasoline in a car engine burns at \(631{ }^{\circ} \mathrm{C},\) while the exhaust temperature (the temperature of the cold reservoir) is \(139^{\circ} \mathrm{C}\) and the outdoor temperature is \(27^{\circ} \mathrm{C}\). Assume that the engine can be treated as a Carnot engine (a gross oversimplification). In an attempt to increase mileage performance, an inventor builds a second engine that functions between the exhaust and outdoor temperatures and uses the exhaust heat to produce additional work. Assume that the inventor's engine can also be treated as a Carnot engine. Determine the ratio of the total work produced by both engines to that produced by the first engine alone.

Short Answer

Expert verified
The ratio of total work to the first engine's work is approximately 1.228.

Step by step solution

01

Convert Temperatures to Kelvin

To use the Carnot efficiency formula, we need the temperatures in Kelvin. Convert each temperature: \( T_1 = 631 + 273 = 904 \, K \), \( T_2 = 139 + 273 = 412 \, K \), and \( T_3 = 27 + 273 = 300 \, K \) for the outdoor temperature.
02

Calculate Efficiency of the First Engine

The efficiency of a Carnot engine is given by \( \eta_1 = 1 - \frac{T_{cold}}{T_{hot}} \). For the first engine, \( \eta_1 = 1 - \frac{412}{904} \approx 0.5446 \).
03

Calculate Efficiency of the Second Engine

Similarly, for the second engine between exhaust and outdoor temperature, \( \eta_2 = 1 - \frac{300}{412} \approx 0.2728 \).
04

Calculate Work for Each Engine

Let \( Q_1 \) be the thermal energy input to the first engine. The useful work output by the first engine is \( W_1 = \eta_1 \cdot Q_1 \). The heat rejected from the first engine is \( Q' = (1 - \eta_1) \cdot Q_1 = 0.4554 \cdot Q_1 \), which acts as the input heat for the second engine. The work output by the second engine is \( W_2 = \eta_2 \cdot Q' = 0.2728 \cdot 0.4554 \cdot Q_1 \).
05

Determine Total Work Produced

Calculate the total work: \( W_{total} = W_1 + W_2 \). Express \( W_2 \) in terms of \( Q_1 \) using the above result: \( W_2 = 0.2728 \cdot 0.4554 \cdot Q_1 \approx 0.1242 \cdot Q_1 \). Thus, \( W_{total} = 0.5446 \cdot Q_1 + 0.1242 \cdot Q_1 = 0.6688 \cdot Q_1 \).
06

Find the Ratio of Total Work to First Engine's Work

Find the ratio \( R = \frac{W_{total}}{W_1} = \frac{0.6688 \cdot Q_1}{0.5446 \cdot Q_1} \approx 1.228 \). This ratio represents the increase in work due to the addition of the second engine.

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It is crucial for understanding processes involving temperature changes, energy flow, and work. In the context of engines, it helps us understand how energy is converted into work, and how to optimize these processes. In this scenario, the Carnot engine, an idealized type of heat engine, helps illustrate these concepts. It perfectly demonstrates the second law of thermodynamics, which states that no engine can be 100% efficient in converting heat into work.

The Carnot engine operates between two heat reservoirs: a high-temperature reservoir (where the engine receives heat) and a low-temperature reservoir (where the engine expels some of this heat). By placing a second engine that uses excess heat from the exhaust of the first engine, we can better utilize this energy according to thermodynamic principles. This highlights how understanding these concepts can help in engineering applications such as improving car engine performance.
Engine Efficiency
Engine efficiency is a measure of how well an engine converts the input energy into useful work. For the Carnot engine, this is calculated using the formula: \[\eta = 1 - \frac{T_{cold}}{T_{hot}}\] where - \(\eta\) is the efficiency,
- \(T_{cold}\) is the temperature of the cold reservoir (in Kelvin), and
- \(T_{hot}\) is the temperature of the hot reservoir (in Kelvin).

This formula shows that efficiency increases as the difference between the hot and cold reservoir temperatures becomes larger.

In the exercise provided, two engines are discussed. The first engine operates between the temperatures of the burned gasoline and the exhaust, while the second engine works between the exhaust and the outdoor temperature. By improving the overall efficiency, the total work produced can exceed that of the initial engine alone.

This approach of using multiple stages or engines can enhance fuel economy and reduce energy waste, showing a practical application of efficient energy conversion.
Kelvin Temperature Conversion
Temperature conversion is key when dealing with thermodynamics, especially as the Kelvin scale is used in scientific calculations. The Kelvin scale starts at absolute zero, which is the theoretical point of no thermal energy. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

For the engine example, temperatures are converted as follows:
- Burned gasoline: \(631^{\circ}C + 273 = 904 \, K\),
- Exhaust: \(139^{\circ}C + 273 = 412 \, K\), - Outdoor: \(27^{\circ}C + 273 = 300 \, K\).

These conversions are vital because the efficiency calculations in thermodynamics require temperature inputs in Kelvin. Using the correct scale ensures precision and accuracy in scientific and engineering calculations, aiding in creating more efficient systems.

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Most popular questions from this chapter

A power plant taps steam superheated by geothermal energy to \(505 \mathrm{K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at \(323 \mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and (b) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

In exercising, a weight lifter loses \(0.150 \mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times\) \(10^{5} \mathrm{J} .\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg},\) find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food (1 nutritional Calorie \(=4186 \mathrm{J}\) ) that must be consumed to replace the loss of internal energy.

A 52-kg mountain climber, starting from rest, climbs a vertical distance of \(730 \mathrm{m}\). At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

On a cold day, 24500 J of heat leaks out of a house. The inside temperature is \(21^{\circ} \mathrm{C},\) and the outside temperature is \(-15^{\circ} \mathrm{C} .\) What is the increase in the entropy of the universe that this heat loss produces?

A monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) is contained within a perfectly insulated cylinder that is fitted with a movable piston. The initial pressure of the gas is \(1.50 \times 10^{5}\) Pa. The piston is pushed so as to compress the gas, with the result that the Kelvin temperature doubles. What is the final pressure of the gas?

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