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A car engine delivers \(8.18 \mathrm{~kJ}\) of work per cycle. (a) Before a tune-up, the efficiency is \(25.0 \% .\) Calculate, per cycle, the heat absorbed from the combustion of fuel and the heat exhausted to the atmosphere. (b) After a tune-up, the efficiency is \(31.0 \%\). What are the new values of the quantities calculated in \((a) ?\)

Short Answer

Expert verified
(a) Before the tune-up: Heat absorbed from the combustion of fuel is 32.72 kJ and the heat exhausted to the atmosphere is 24.54 kJ. (b) After the tune-up: Heat absorbed from the combustion of fuel is 26.39 kJ and the heat exhausted to the atmosphere is 18.21 kJ.

Step by step solution

01

Calculate the heat energy absorbed (Qin) before the tune-up

The formula for efficiency (\(\eta\)) is \(\eta = \frac{W}{Q_{in}}\), where \(W\) is work and \(Q_{in}\) is the heat energy absorbed. From this, we can rearrange to find \(Q_{in}\) as \(Q_{in} = \frac{W}{\eta}\). Substituting \(W = 8.18 kJ\) and \(\eta = 0.25\) gives \(Q_{in} = \frac{8.18 kJ}{0.25} '= 32.72 kJ \)
02

Calculate the heat energy exhausted (Qout) before the tune-up

The formula for the heat energy exhausted is \(Q_{out} = Q_{in} - W\). Substituting \(Q_{in} = 32.72 kJ\) and \(W = 8.18 kJ\) gives \(Q_{out} = 32.72 kJ - 8.18 kJ = 24.54 kJ\)
03

Calculate the heat energy absorbed (Qin) after the tune-up

After the tune-up, the efficiency is \(\eta = 0.31\). We can use the formula for \(Q_{in}\) from step 1 to find the new \(Q_{in}\). Substituting \(W = 8.18 kJ\) and \(\eta = 0.31\) gives \(Q_{in} '= \frac{8.18 kJ}{0.31} '= 26.39 kJ \)
04

Calculate the heat energy exhausted (Qout) after the tune-up

Using the formula for \(Q_{out}\) from step 2, we can find the new \(Q_{out}\). Substituting \(Q_{in} '= 26.39 kJ\) and \(W = 8.18 kJ\) gives \(Q_{out} '= 26.39 kJ - 8.18 kJ = 18.21 kJ\)

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Key Concepts

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

Work-Energy Principle
Understanding the work-energy principle is key when studying the mechanics of how engines operate. Essentially, this principle states that work, which is the effort over a distance, can be converted into energy, and vice versa. In the context of an engine, when fuel combusts, chemical energy is converted into thermal energy, and then into mechanical work, propelling the vehicle forward.

Using the work-energy principle, we can calculate the amount of work an engine can perform based on the heat energy absorbed from the fuel. Mathematically, if an engine delivers a work output of \(8.18 \mathrm{~kJ}\), we apply the work-energy principle to relate this output to the thermal energy input derived from the combustion process.
Heat Transfer
Heat transfer in an engine context refers to the movement of thermal energy from one place to another. In engines, this involves the transfer of heat from the combustion of fuel to do work and also to the surrounding environment as waste heat due to inefficiency.

Applications of heat transfer are crucial to understanding engine efficiency. Before a tune-up, if an engine has a 25.0% efficiency, it means that only a quarter of the heat energy from the fuel is converted to work, while the rest is exhausted or dissipated into the atmosphere as 'waste heat'.
Engine Tune-Up
An engine tune-up is a series of adjustments and calibrations made to an engine to improve its performance and efficiency. During a tune-up, components such as spark plugs, air filters, and fuel injectors are inspected and serviced to ensure that the engine runs smoothly and that fuel combustion is optimized.

Improvements from a tune-up can lead to increased thermal efficiency. For example, after a tune-up, the efficiency of the car engine in our exercise increased from 25.0% to 31.0%. This means that more of the fuel's heat energy gets converted into work and less is wasted, signifying that tune-ups play a critical role in engine performance and efficiency.
Conservation of Energy
The conservation of energy is a fundamental concept in physics which states that energy cannot be created or destroyed, only transformed from one form to another. In any system, including engines, the total energy remains constant. When an engine operates, the chemical energy stored in the fuel is converted into mechanical energy and thermal energy.

Considering conservation of energy enables us to understand that the work produced by the engine plus the heat exhausted to the atmosphere equals the total heat absorbed from the fuel combustion. It's crucial in calculating engine efficiency, as it helps track where energy is lost and how it can be utilized more effectively during operation.

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Most popular questions from this chapter

An inventor claims to have created a heat pump that draws heat from a lake at \(3.0^{\circ} \mathrm{C}\) and delivers heat at a rate of \(20 \mathrm{~kW}\) to a building at \(35^{\circ} \mathrm{C}\), while using only \(1.9 \mathrm{~kW}\) of electrical power. How would you judge the claim?

In a refrigerator the low-temperature coils are at a temperature of \(-13^{\circ} \mathrm{C}\) and the compressed gas in the condenser has a temperature of \(25^{\circ} \mathrm{C}\). Find the coefficient of performance of a Carnot refrigerator operating between these temperatures.

(a) A Carnot engine operates between a hot reservoir at \(322 \mathrm{~K}\) and a cold reservoir at \(258 \mathrm{~K}\). If it absorbs \(568 \mathrm{~J}\) of heat per cycle at the hot reservoir, how much work per cycle does it deliver? (b) If the same engine, working in reverse, functions as a refrigerator between the same two reservoirs, how much work per cycle must be supplied to transfer \(1230 \mathrm{~J}\) of heat from the cold reservoir?

Four moles of an ideal gas are caused to expand from a volume \(V_{1}\) to a volume \(V_{2}=3.45 V_{1} .(a)\) If the expansion is isothermal at the temperature \(T=410 \mathrm{~K}\), find the work done on the expanding gas. ( \(b\) ) Find the change in entropy, if any. ( \(c\) ) If the expansion is reversibly adiabatic instead of isothermal, what is the entropy change?

To make some ice, a freezer extracts \(185 \mathrm{~kJ}\) of heat at \(-12.0^{\circ} \mathrm{C} .\) The freezer has a coefficient of performance of 5.70. The room temperature is \(26.0^{\circ} \mathrm{C}\). (a) How much heat is delivered to the room? ( \(b\) ) How much work is required to run the freezer?

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