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Due to a tune-up, the efficiency of an automobile engine increases by \(5.0 \% .\) For an input heat of \(1300 \mathrm{J},\) how much more work does the engine produce after the tune-up than before?

Short Answer

Expert verified
The engine produces 65 J more work after the tune-up.

Step by step solution

01

Understanding Efficiency

Efficiency (\(\eta\)) is the ratio of the work output (\(W\)) to the heat input (\(Q_{in}\)), expressed as \(\eta = \frac{W}{Q_{in}}\). Before the tune-up, let's denote the initial efficiency as \(\eta_0\), and after the tune-up, the increased efficiency as \(\eta_1 = \eta_0 + 0.05\).
02

Express Initial and Final Work

Before the tune-up, the work produced is \(W_0 = \eta_0 \times Q_{in}\). After the tune-up, the work produced is \(W_1 = \eta_1 \times Q_{in}\). Since \(\eta_1 = \eta_0 + 0.05\), the expression for \(W_1\) becomes \(W_1 = (\eta_0 + 0.05) \times Q_{in}\).
03

Calculate the Increase in Work

The increase in work \(\Delta W\) is given by \(\Delta W = W_1 - W_0\). Substituting the expressions for \(W_1\) and \(W_0\), we have \(\Delta W = ((\eta_0 + 0.05) \times Q_{in}) - (\eta_0 \times Q_{in})\). Simplifying gives \(\Delta W = 0.05 \times Q_{in}\).
04

Calculate the Additional Work Produced

Since the input heat \(Q_{in}\) is given as 1300 J, substitute this into the equation: \(\Delta W = 0.05 \times 1300\). This calculation results in \(\Delta W = 65 \text{ J}\).

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Key Concepts

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

Work Output
When discussing automobile engine efficiency, it's important to understand the concept of work output. Work output is the useful energy produced by the engine from the fuel it consumes. This determines how efficient the engine is in converting energy from the chemical form (fuel) into mechanical energy (movement).

To calculate work output, we use the formula:

\[ W = ext{Efficiency} imes ext{Heat Input} \]

Where:
  • \(W\) is the work output,
  • Efficiency (\(\eta\)) is the ratio of energy converted to work,
  • Heat Input (\(Q_{in}\)) is the total energy available from burning the fuel.
In the case of a more efficient engine, the same amount of heat input results in more work output. Adjustments or tune-ups can improve engine efficiency, leading to increased work output even when the heat input remains unchanged.
Heat Input
Heat input refers to the amount of energy supplied to the engine from fuel combustion. It is the energy potential available to be converted into effective work output. Heat input can vary based on the fuel type, amount, and combustion efficiency.

Understanding heat input is crucial because it sets the upper limit on how much work can be done by the engine. If you know the heat input and efficiency, you can determine how much work is achieved.

Using the formula:

\[ ext{Work Output} = ext{Efficiency} imes ext{Heat Input} \]

With higher heat input, there's more energy available for work output. However, this also depends on how efficiently the engine converts this energy. Efficient energy conversion means getting more work for the same amount of heat input.
Energy Conversion
Energy conversion in an automobile engine involves transforming the chemical energy stored in fuel into mechanical energy, which moves the vehicle. This conversion is pivotal because it depicts how well the engine utilizes the energy available to perform work.

In terms of efficiency, not all the input energy from the fuel is converted to mechanical work. Some energy is always lost as waste heat due to factors like friction and incomplete combustion. This is why increasing conversion efficiency is critical for better performance.

The energy conversion process can be seen through this equation:

\[ ext{Efficiency} = \frac{ ext{Work Output} }{ ext{Heat Input} } \]

This equation shows that with a higher efficiency, a greater portion of the heat input translates into work output. Advances in engine design focus on maximizing energy conversion efficiency.
Efficiency Increase
An efficiency increase in an automobile engine means the engine becomes better at converting heat input to useful work. From a user perspective, this can mean more power for the same amount of fuel, or the same power with less fuel consumed.

A tune-up can help achieve this efficiency increase, often involving adjustments such as optimizing the air-fuel mixture, improving ignition timing, or reducing friction. These modifications improve how the engine uses available energy.

For example, if an engine's efficiency increases by 5%, it means that for the same amount of fuel, the engine will produce 5% more work output. The realization can be calculated via:

\[ ext{Increase in Work} = ext{Efficiency Increase} imes ext{Heat Input} \]

This efficiency increase results in tangible benefits such as reduced fuel costs and enhanced vehicle performance. Efficient engines contribute to better fuel economy and lower emissions, fostering sustainability.

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Most popular questions from this chapter

An engine has an efficiency \(e_{1}\). The engine takes input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) from a hot reservoir and delivers work of magnitude \(\left|W_{1}\right| .\) The heat rejected by this engine is used as input heat for a second engine, which has an efficiency \(e_{2}\) and delivers work of magnitude \(\left|W_{2}\right|\). The overall efficiency of this two- engine device is the magnitude of the total work delivered \(\left(\left|W_{1}\right|+\left|W_{2}\right|\right)\) divided by the magnitude \(\left|Q_{\mathrm{H}}\right|\) of the input heat. Find an expression for the overall efficiency \(e\) in terms of \(e_{1}\) and \(e_{2}\).

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.

A Carnot engine operates between temperatures of 650 and \(350 \mathrm{K} .\) To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by \(40 \mathrm{K}\) or to lower the temperature of the cold reservoir by \(40 \mathrm{K}\). Which change gives the greater improvement? Justify your answer by calculating the efficiency in each case.

A system gains \(1500 \mathrm{J}\) of heat, while the internal energy of the system increases by \(4500 \mathrm{J}\) and the volume decreases by \(0.010 \mathrm{m}^{3} .\) Assume that the pressure is constant and find its value.

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11. (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times\) \(10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.

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