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(III) A typical compact car experiences a total drag force of about 350 N at 55 mi/h. If this car gets 32 miles per gallon of gasoline at this speed, and a liter of gasoline (1 gal \(=\) 3.8 L) releases about \(3.2 \times 10^7\) J when burned, what is the car's efficiency?

Short Answer

Expert verified
The car's efficiency is 56.32%.

Step by step solution

01

Convert Speed to Meters per Second

First, convert the speed from miles per hour to meters per second. We know that 1 mile = 1609.34 meters and 1 hour = 3600 seconds, thus: \[ 55 \text{ mi/h} = 55 \times 1609.34 \text{ meters} = 88413.7 \text{ meters}. \]\[ \frac{88413.7 \text{ meters}}{3600 \text{ seconds}} = 24.56 \text{ m/s}. \]So, 55 mi/h is equivalent to 24.56 m/s.
02

Convert Miles per Gallon to Kilometers per Liter

Now we convert the fuel efficiency from miles per gallon to kilometers per liter using the conversion factors (1 mile = 1.60934 km and 1 gallon = 3.8 liters):\[ 32 \text{ miles/gallon} = 32 \times 1.60934 \text{ kilometers} = 51.4988 \text{ kilometers}. \]\[ \frac{51.4988 \text{ kilometers}}{3.8 \text{ liters}} = 13.55 \text{ km/L}. \]So, the car's fuel efficiency is 13.55 km/L.
03

Calculate Work Done per Liter of Gasoline

The work done per liter of gasoline when driving at constant speed can be determined using the relation between force, distance, and energy:\[ F = 350 \text{ N}, \quad \text{Fuel efficiency} = 13.55 \text{ km/L}. \]Distance per liter: 13.55 km = 13550 meters.The work done (or energy used) to overcome the drag while traveling 13.55 km is: \[ \text{Work} = F \times \text{Distance} = 350 \times 13550 = 4742500 \text{ J/L}. \]
04

Calculate Energy Released per Liter of Gasoline

Given that the energy released by burning 1 gallon of gasoline is approximately \(3.2 \times 10^7\) J, we first convert this to per liter:\[ \text{Energy released per liter} = \frac{3.2 \times 10^7}{3.8} = 8421052.6 \text{ J/L}. \]
05

Calculate Efficiency

Finally, efficiency can be calculated as the ratio of useful work performed (output) to the total energy produced (input): \[ \text{Efficiency} = \frac{\text{Work}}{\text{Energy released}} = \frac{4742500}{8421052.6}. \]Therefore, the efficiency is:\[ \text{Efficiency} = 0.5632 \text{ or } 56.32\% \].

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Key Concepts

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

Understanding Drag Force
In the context of car efficiency, drag force plays a crucial role. Drag force is a resistance force exerted by air or any fluid that opposes the motion of an object. For cars, drag force impacts how much energy they require to travel at a certain speed.
The larger the drag force, the more energy a car needs to maintain its speed. This force can depend on several factors like the car's shape, speed, and surface area facing the oncoming air.
In our example, the car encounters a drag force of 350 Newtons (N). This resistance is an energy loss that must be overcome by the vehicle's engine, thus affecting the car's fuel efficiency.
Explaining Unit Conversion
Unit conversion is a necessary step in many physics problems, especially when dealing with car efficiency calculations. Different systems use different units of measurement, which can complicate direct comparison and calculations.
For instance, speed is often measured in miles per hour in the U.S. but needs to be converted to meters per second for scientific calculations. Similarly, fuel efficiency might be provided in miles per gallon but needs conversion to kilometers per liter.
To convert these units accurately:
  • 1 mile equals approximately 1609.34 meters.
  • 1 hour equals 3600 seconds, helping change miles per hour to meters per second.
  • 1 mile equates to 1.60934 kilometers, and 1 gallon is the same as 3.8 liters for converting fuel efficiency.
These conversions ensure consistency in calculations and allow for precise determination of car performance metrics.
Understanding Fuel Efficiency
Fuel efficiency indicates how far a car can travel per unit of fuel. It's a critical factor when assessing the energy performance of a vehicle. Fuel efficiency tells us how economically a vehicle uses fuel, essential for cost saving and energy conservation.
In our example, the car's fuel efficiency is originally given as 32 miles per gallon. Using unit conversions, this was transformed into 13.55 kilometers per liter (km/L).
Higher fuel efficiency means a car can travel more kilometers per liter of fuel. This ultimately reduces the frequency of refueling, leading to cost savings and reduced environmental impact.
Understanding a vehicle's fuel efficiency is key to optimizing its energy consumption and performance on the road.
Work and Energy Concepts
The principles of work and energy are foundational to physics and play a significant part in vehicle efficiency analysis. Work is defined as the product of force and distance, expressed in joules (J) when using the metric system.
In the context of our car moving at a constant speed, the work done is essentially the energy required to overcome the drag force. Given our car's drag force and its travel distance per liter of fuel, we calculate the work done per liter of fuel.
  • Work can be calculated as force multiplied by the distance the car travels per liter (Energy used in Joules).
Meanwhile, energy refers to the capacity to do work. The energy quantity is often derived from fuel, released as the chemical energy from gasoline combustion.
In our example, burning 1 liter of gasoline releases approximately 8421052.6 J of energy. Efficiency measures become clear by comparing the work done against this energy release, reflecting how effectively the energy potential of fuel is used to perform work in overcoming drag.

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Most popular questions from this chapter

Refrigeration units can be rated in "tons." A 1-ton air conditioning system can remove sufficient energy to freeze 1 ton (2000 pounds \(=\) 909 kg) of 0\(^\circ\)C water into 0\(^\circ\)C ice in one 24-h day. If, on a 35\(^\circ\)C day, the interior of a house is maintained at 22\(^\circ\)C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs \$0.10 per kWh and that the unit's coefficient of performance is 18\(\%\) that of an ideal refrigerator. 1 kWh \(= 3.60 \times 10^6\) J.

A particular car does work at the rate of about 7.0 kJ/s when traveling at a steady 21.8 m/s along a level road. This is the work done against friction. The car can travel 17 km on 1.0 L of gasoline at this speed (about ). What is the minimum value for \(T_H\) if \(T_L\) is 25\(^\circ\)C? The energy available from 1.0 L of gas is \(3.2 \times 10^7\) J.

(II) A nuclear power plant operates at 65\(\%\) of its maximum theoretical (Carnot) efficiency between temperatures of 660\(^\circ\)C and 330\(^\circ\)C. If the plant produces electric energy at the rate of 1.4GW, how much exhaust heat is discharged per hour?

(II) Which will improve the efficiency of a Carnot engine more: a 10 C\(^\circ\) increase in the high-temperature reservoir, or a 10 C\(^\circ\) decrease in the low-temperature reservoir? Give detailed results. Can you state a generalization?

(II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a \(PV\) diagram, including numbers and labels for the axes.

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