/*! 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 54 A car engine burns 5 kg fuel (eq... [FREE SOLUTION] | 91Ó°ÊÓ

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A car engine burns 5 kg fuel (equivalent to addition of \(Q_{H}\) ) at \(1500 \mathrm{~K}\) and rejects energy to the radiator and the exhaust at an average temperature of \(750 \mathrm{~K}\). If the fuel provides \(40000 \mathrm{~kJ} / \mathrm{kg}\), what is the maximum amount of work the engine can provide?

Short Answer

Expert verified
The maximum work the engine can provide is 100,000 kJ.

Step by step solution

01

Understanding the Problem

We need to calculate the maximum work a car engine can produce given the temperature and energy conditions. This involves using the concept of a Carnot engine, which provides the maximum possible efficiency between two temperatures.
02

Identify the Given Values

The given values include the high temperature of the fuel (\(T_H = 1500 \ \text{K}\)), the low temperature of the exhaust and radiator (\(T_C = 750 \ \text{K}\)), the energy provided by the fuel per kilogram (\(q_H = 40000 \ \text{kJ/kg}\)), and the mass of the fuel (\(m = 5 \ \text{kg}\)).
03

Calculate the Total Heat Input (\(Q_H\))

Multiply the energy provided by the fuel per kilogram by the mass of the fuel to find the total energy input:\[ Q_H = q_H \times m = 40000 \ \text{kJ/kg} \times 5 \ \text{kg} = 200000 \ \text{kJ} \]
04

Determine the Carnot Efficiency Formula

The Carnot efficiency formula is given by:\[ \eta = 1 - \frac{T_C}{T_H} \]
05

Calculate the Carnot Efficiency

Substitute the given temperatures into the Carnot efficiency formula to find the maximum possible efficiency:\[ \eta = 1 - \frac{750}{1500} = 1 - 0.5 = 0.5 \]
06

Calculate the Maximum Work (\(W_{max}\))

Multiply the Carnot efficiency by the total heat input to find the maximum work the engine can provide:\[ W_{max} = \eta \times Q_H = 0.5 \times 200000 \ \text{kJ} = 100000 \ \text{kJ} \]
07

Conclusion

The maximum amount of work that the engine can provide thanks to the energy it receives from burning fuel is determined by the Carnot efficiency of the system.

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Key Concepts

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

Carnot Efficiency
In thermodynamics, the concept of Carnot efficiency is fundamental when analyzing the maximum possible efficiency that any heat engine can achieve. Carnot efficiency represents the highest efficiency any cycle can achieve when operating between two temperatures, according to the second law of thermodynamics. This theoretical efficiency is always based on reversible processes.
For a heat engine operating between a hot reservoir with temperature \(T_H\) and a cold reservoir with temperature \(T_C\), the Carnot efficiency \(\eta\) is given by the formula:
  • \(\eta = 1 - \frac{T_C}{T_H} \)
The temperatures \(T_H\) and \(T_C\) must be in absolute units like Kelvin. For our car engine example, the Carnot efficiency can be calculated by substituting the given temperatures: \(T_H = 1500 \ \text{K}\) and \(T_C = 750 \ \text{K}\). This results in:
  • \(\eta = 1 - \frac{750}{1500} = 0.5 \)
This means that even under perfect conditions, the engine can convert only 50% of the heat energy it absorbs into work.
Heat Engine
A heat engine is a device that converts thermal energy into mechanical work by exploiting the temperature difference between a hot and a cold reservoir. This principle forms the basis of many practical engines and power-generation systems, including car engines. The operating principle revolves around utilizing heat flow naturally from hot to cold areas to perform work.
In the context of our problem, the car's engine acts as a heat engine by taking in the thermal energy from burning fuel and then converting a portion of that energy into mechanical work, while the rest is discharged to the environment.
Heat engines are evaluated based on their efficiency, which is the ratio of work output to heat input. By comparing real engine efficiencies to Carnot efficiency, engineers can assess how close their device performs compared to the ideal scenario.
Work Calculation
In thermodynamics, the work done by a heat engine is closely linked to its efficiency and the heat input it receives. The maximum work output, \(W_{max}\), can be determined by multiplying the Carnot efficiency \(\eta\) by the total heat input \(Q_H\). Showcasing all inputs into this formula helps in clear understanding and calculation:
  • Given: \(\eta = 0.5\), \(Q_H = 200000 \ \text{kJ}\)
  • Maximum Work \(W_{max} = \eta \times Q_H\)
  • Substituting the values: \(W_{max} = 0.5 \times 200000 \ \text{kJ} = 100000 \ \text{kJ}\)
This calculation tells us the maximum amount of work the engine can output under ideal conditions, relying on the Carnot efficiency for its upper limit. Real engines, however, usually have lower work output due to practical limitations such as friction and non-reversible processes.
Temperature Difference
The temperature difference between the reservoirs is crucial in determining both the efficiency and work output of a heat engine. A larger temperature difference typically allows for a higher potential for work extraction. The temperatures involved in this process are of particular importance:
  • \(T_H\) is the temperature of the hot reservoir.
  • \(T_C\) is the temperature of the cold reservoir.
In the car engine scenario, \(T_H = 1500\ \text{K}\), and \(T_C = 750\ \text{K}\) create a difference, or gradient, that determines how effectively heat can be transformed into work.
By understanding this, students can see why increasing the temperature of the heat source or decreasing the temperature of the heat sink can improve the theoretical efficiency of any real-world heat engine.

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Most popular questions from this chapter

A cyclic machine, shown in Fig. P5.73, receives \(325 \mathrm{~kJ}\) from a \(1000-\mathrm{K}\) energy reservoir. It rejects \(125 \mathrm{~kJ}\) to a \(400-\mathrm{K}\) energy reservoir, and the cycle produces \(200 \mathrm{~kJ}\) of work as output. Is this cycle reversible, irreversible, or impossible?

A steam power plant has \(1200 \mathrm{~F}\) in the boiler, \(630 \mathrm{Btu} / \mathrm{s}\) work out of the turbine, \(900 \mathrm{Btu} / \mathrm{s}\) is taken out at \(100 \mathrm{~F}\) in the condenser, and the pump work is 30 Btu/s. Find the plant's thermal efficiency. Assuming the same pump work and heat transfer to the boiler, what is the turbine power if the plant is running in a Carnot cycle?

The management at a large factory cannot decide which of two fuels to purchase. The selected fuel will be used in a heat engine operating between the fuel-burning temperature and a low-exhaust temperature. Fuel A burns at \(2200 \mathrm{~K}\) and exhausts at \(450 \mathrm{~K},\) delivering \(30000 \mathrm{~kJ} / \mathrm{kg},\) and costs \(\$ 1.50 / \mathrm{kg} .\) Fuel B burns at \(1200 \mathrm{~K}\) and exhausts at \(350 \mathrm{~K},\) delivering \(40000 \mathrm{~kJ} / \mathrm{kg},\) and costs \(\$ 1.30 / \mathrm{kg} .\) Which fuel would you buy and why?

A \(4 \mathrm{~L}\) jug of milk at \(25^{\circ} \mathrm{C}\) is placed in your refrigerator, where it is cooled down to \(5^{\circ} \mathrm{C}\). The high temperature in the Carnot refrigeration cycle is \(45^{\circ} \mathrm{C}\), the low temperature is \(-5^{\circ} \mathrm{C},\) and the properties of milk are the same as those of liquid water. Find the amount of energy that must be removed from the milk and the additional work needed to drive the refrigerator.

A room is heated with a 1500 -W electric heater. How much power can be saved if a heat pump with a COP of 2.5 is used instead?

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