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An ideal heat engine operates between \(405 \mathrm{~K}\) and \(305 \mathrm{~K}\). Given that it receives \(16670 \mathrm{~J}\) of heat from the high-temperature source during each cycle, how much work does it do? How much heat does it exhaust?

Short Answer

Expert verified
Work done is 4114 J; heat exhausted is 12556 J.

Step by step solution

01

Identify temperatures

Identify the high-temperature ( T_h ) and low-temperature ( T_c ) reservoirs in Kelvin. In this problem, T_h = 405 ext{ K} and T_c = 305 ext{ K} .
02

Calculate efficiency of the heat engine

The efficiency d_{ ext{Carnot}} of a Carnot engine is given by the formula:\( de_{ ext{Carnot}} = 1 - \frac{T_c}{T_h}\). Substitute T_c = 305 ext{ K} and T_h = 405 ext{ K} to find the efficiency: \( de_{ ext{Carnot}} = 1 - \frac{305}{405} \). Perform the calculation to get the efficiency, de_{ ext{Carnot}} \approx 0.2469 (or 24.69%).
03

Calculate work done by the heat engine

The work W done by the engine is found using: \( W = de_{ ext{Carnot}} imes Q_h \), where Q_h = 16670 ext{ J}. Substitute the values to get \( W = 0.2469 imes 16670 \). Calculate to find the work done, W \approx 4114.023 ext{ J} simply round to 4114 ext{ J}.
04

Calculate the heat exhausted by the engine

The heat exhausted Q_c by the engine is given by the formula: \( Q_c = Q_h - W \). Now substitute the values Q_h = 16670 ext{ J} and W = 4114 ext{ J}. \( Q_c = 16670 - 4114 \). Calculate to find the heat exhausted, Q_c \approx 12556 ext{ J}.

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Key Concepts

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

Ideal Heat Engine
The idea of an ideal heat engine is fundamental in thermodynamics. An ideal heat engine, such as a Carnot engine, operates under a reversible process that maximizes efficiency. This means there is no energy lost to friction or any other inefficiencies typically found in real engines. In a Carnot cycle, the engine absorbs heat from a high-temperature source, converts some of it into work, and exhausts the remaining heat to a low-temperature sink. This concept serves as a benchmark, helping us understand the upper limits of efficiency an engine can achieve. In reality, no engine can be perfectly ideal due to inevitable material limitations and frictional losses.
Thermodynamic Efficiency
Thermodynamic efficiency is a key aspect of heat engines and signifies how well an engine converts input heat energy into work. It is calculated using the formula for Carnot efficiency: \[ \eta_{Carnot} = 1 - \frac{T_c}{T_h} \]where \(T_h\) is the high temperature and \(T_c\) is the low temperature measured in Kelvin.
  • The closer the temperatures, the lower the efficiency, as less heat is converted to work.
  • The efficiency of a Carnot engine is the theoretical maximum efficiency since it assumes no energy losses due to friction or other non-ideal circumstances.
In our example, with high and low temperatures at 405 K and 305 K, respectively, the efficiency is approximately 24.69%. This means about a quarter of the heat energy can be potentially converted into work under ideal conditions.
Work Done Calculation
The work done by an ideal heat engine during each cycle is the product of the engine's efficiency and the heat absorbed from the high-temperature source. Using the formula:\[ W = \eta_{Carnot} \times Q_h \]we calculate the work, where \(Q_h\) is the heat received, and \(\eta_{Carnot}\) is the efficiency we've previously calculated. For this engine, receiving 16,670 J of heat, the work produced is approximately 4,114 J.This work represents the useful energy that can be applied to perform tasks, such as spinning a turbine or driving mechanical processes. Understanding how much work is generated helps in evaluating the engine's performance and its efficiency in energy conversion.
Heat Energy Transfer
Heat energy transfer involves the absorption of heat by the engine from the high-temperature reservoir and the rejection of waste heat to the low-temperature reservoir.In the Carnot cycle, the heat transfer follows the principle of conservation of energy, stating that the total energy input minus the energy output used as work results in the heat exhausted. For our problem:\[ Q_c = Q_h - W \]where \(Q_c\) is the heat exhausted, \(Q_h\) is the heat input, and \(W\) is the work done. In this example, after subtracting the work done (4,114 J) from the received heat (16,670 J), the heat exhausted is found to be 12,556 J.This transfer is critical to understanding how energy flows within the engine, emphasizing the importance of temperature differences in facilitating this exchange.

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Most popular questions from this chapter

If a person does \(8.00 \mathrm{~h}\) of moderate physical labor "burning" 400 kcal/h, by how much does his or her internal energy change as a result?

In a certain process, \(8.00\) kcal of heat is furnished to the system while the system does \(6.00 \mathrm{~kJ}\) of work. By how much does the internal energy of the system change during the process? Here \(8.00\) kcal is heat-in and \(6.00 \mathrm{~kJ}\) is work-out, both of which are positive. Consequently, \(\Delta Q=(8000\) cal \()(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}\) and \(\Delta W=6.00 \mathrm{~kJ}\) Therefore, from the First Law \(\Delta Q=\Delta U+\Delta W\), $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

The specific heat of water is \(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). By how many joules does the internal energy of \(50 \mathrm{~g}\) of water change as it is heated from \(21^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\) ? Assume that the expansion of the water is negligible. The heat added to raise the temperature of the water is \(\Delta Q=c m \Delta T=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.050 \mathrm{~kg})\left(16^{\circ} \mathrm{C}\right)=3.4 \times 10^{3} \mathrm{~J}\) Notice that \(\Delta T\) in Celsius is equal to \(\Delta T\) in kelvin. If we ignore the slight expansion of the water, no work was done on the surroundings and so \(\Delta W=0\). Then, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=3.4 \mathrm{~kJ} $$

How much external work is done by an ideal gas in expanding from a volume of \(3.0\) liters to a volume of \(30.0\) liters against a constant pressure of \(2.0\) atm?

A cylinder of ideal gas is closed by an \(8.00\) kg movable piston \(\left(\right.\) area \(\left.=60.0 \mathrm{~cm}^{2}\right)\) as illustrated in \(\underline{\text { Fig. } 20-5}\). Atmospheric pressure is \(100 \mathrm{kPa}\). When the gas is heated from \(30.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\), the piston rises \(20.0 \mathrm{~cm}\). The piston is then fastened in place, and the gas is cooled back to \(30.0^{\circ} \mathrm{C}\). Calling \(\Delta Q_{1}\) the heat added to the gas in the heating process, and \(\Delta Q_{2}\) the heat lost during cooling, find the difference between \(\Delta Q_{1}\) and \(\Delta Q_{2}\). During the heating process, the internal energy changed by \(\Delta U_{1}\), and an amount of work \(\Delta W_{1}\) was done. The absolute pressure of the gas was $$P=\frac{m g}{A}+P_{A}$$ $$P=\frac{(8.00)(9.81) \mathrm{N}}{60.0 \times 10^{-4} \mathrm{~m}^{2}}+1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.13 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$$ Therefore, $$\begin{aligned} \Delta Q_{1} &=\Delta U_{1}+\Delta W_{1}=\Delta U_{1}+P \Delta V \\ &=\Delta U_{1}+\left(1.13 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.200 \times 60.0 \times 10^{-4} \mathrm{~m}^{3}\right)=\Delta U_{1}+136 \mathrm{~J} \end{aligned}$$ During the cooling process, \(\Delta W=0\) and so (since \(\Delta Q_{2}\) is heat lost) $$-\Delta Q_{2}=\Delta U_{2}$$ But the ideal gas returns to its original temperature, and so its internal energy is the same as at the start. Therefore \(\Delta U_{2}=-\Delta U_{1}\), or \(\Delta Q_{2}=\Delta U_{1} .\) It follows that \(\Delta Q_{1}\) exceeds \(\Delta Q_{2}\) by \(136 \mathrm{~J}=32.5\) cal.

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