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Chapter 22: Problem 6

Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum \(\left(660^{\circ} \mathrm{C}\right)\) and the other a block of solid mercury \(\left(-38.9^{\circ} \mathrm{C}\right)\) . The engine runs by freezing 1.00 \(\mathrm{g}\) of aluminum and melting 15.0 \(\mathrm{g}\) of mercury during each cycle. The heat of fusion of aluminum is \(3.97 \times 10^{5} \mathrm{J} / \mathrm{kg} ;\) the heat of fusion of mercury is \(1.18 \times 10^{4} \mathrm{J} / \mathrm{kg}\) . What is the efficiency of this engine?

Short Answer

Expert verified
The efficiency of the engine is approximately 92%.

Step by step solution

01

Convert mass to kilograms

First, convert the masses of aluminum and mercury from grams to kilograms because the heat of fusion values are given in joules per kilogram.
02

Calculate the heat absorbed (Q_hot)

Calculate the heat absorbed by the engine from the molten aluminum using the formula \(Q_{\text{hot}} = m_{\text{Al}} \times L_{\text{Al}}\), where \(m_{\text{Al}}\) is the mass of aluminum in kilograms and \(L_{\text{Al}}\) is the heat of fusion of aluminum.
03

Calculate the heat expelled (Q_cold)

Calculate the heat expelled by the engine to the block of solid mercury using the formula \(Q_{\text{cold}} = m_{\text{Hg}} \times L_{\text{Hg}}\), where \(m_{\text{Hg}}\) is the mass of mercury in kilograms and \(L_{\text{Hg}}\) is the heat of fusion of mercury.
04

Calculate the work done (W)

The work done by the engine is the difference between the heat absorbed and the heat expelled, given by \(W = Q_{\text{hot}} - Q_{\text{cold}}\).
05

Calculate the efficiency

The efficiency (e) of the engine is given by the ratio of the work done to the heat absorbed, \(e = \frac{W}{Q_{\text{hot}}}\). Convert the efficiency to a percentage by multiplying the result by 100%.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics concerned with the relationships between heat and other forms of energy. In particular, it describes how thermal energy is converted to and from other forms of energy and how it affects matter. A heat engine, which is the focal point of the exercise, is a classic example of thermodynamics in action. It operates between two temperature reservoirs, absorbing heat energy (\(Q_{hot}\)) from a hot source, converting some of it to work (\(W\)), and expelling the remaining energy (\(Q_{cold}\)) to a colder sink.

The performance of such an engine is measured by its efficiency, which is the ratio of work extracted from the engine to the heat energy absorbed from the hot source. The higher the efficiency, the more effective the engine is in converting heat into work. The second law of thermodynamics stipulates that no engine can be 100% efficient when the waste heat (\(Q_{cold}\)) must be discarded into a cooler environment.
Heat of Fusion
The heat of fusion is a specific type of latent heat necessary to change the state of a substance from solid to liquid at constant temperature and pressure. It's an important concept in phase transitions of matter. In the context of the exercise, the heat of fusion lets us calculate how much energy is needed to melt a given mass of mercury (\(L_{Hg}\)) and similarly, how much energy is needed to freeze aluminum (\(L_{Al}\)).

To find the relevant energies involved, we use the masses of the substances in kilograms and multiply by their respective heats of fusion. Since energy is required to melt the mercury and released when aluminum freezes, the heat of fusion values give us critical information needed to calculate the heat absorbed and expelled by the engine鈥攁 key step in determining the engine's efficiency.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In the heat engine cycle, energy conservation is exemplified by the conversion of thermal energy into mechanical work and vice versa. The engine absorbs heat (\(Q_{hot}\)) from the molten aluminum, does work (\(W\)), and the rest of the energy is expelled as heat to the mercury (\(Q_{cold}\)).

In the solution, Step 4 highlights the application of energy conservation where the work done by the engine is the difference between the energy absorbed from the hot reservoir and the energy expelled to the cold reservoir. Mathematically, this is presented as: \[W = Q_{hot} - Q_{cold}\] This equation simplifies the complex interactions of a heat engine into a clear representation of energy flow, which is essential for calculating the engine鈥檚 efficiency.

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Most popular questions from this chapter

Calculate the change in entropy of 250 \(\mathrm{g}\) of water heated slowly from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} .\) (Suggestion: Note that \(d Q=m c d T . )\)

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperature of 720 K. It expands isothermally to point B, and then expands adiabatically to point C where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes and temperatures as you fill in the following table: $$\begin{array}{|c|c|c|c|}\hline & {P} & {V} & {T} \\ \hline A & {1400 \mathrm{kPa}} & {10.0 \mathrm{L}} & {720 \mathrm{K}} \\ \hline B & {} & {} \\\ \hline C & {} & {24.0 \mathrm{L}} \\ \hline D & {} & {15.0 \mathrm{L}} & {} \\ \hline\end{array}$$ (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps \(A \rightarrow B, B \rightarrow C, C \rightarrow D,\) and \(D \rightarrow A .\) (c) Calculate the efficiency \(W_{\text { net }} / Q_{h}\) . Show that it is equal to \(1-T_{C} / T_{A}\) the Carnot efficiency.

Two identically constructed objects, surrounded by thermal insulation, are used as energy reservoirs for a Carnot engine. The finite reservoirs both have mass \(m\) and specific heat \(c\) . They start out at temperatures \(T_{h}\) and \(T_{c}\) , where \(T_{h}>T_{c} .\) (a) Show that the engine will stop working when the final temperature of each object is \(\left(T_{h} T_{c}\right)^{1 / 2}\) (b) Show that the total work done by the Carnot engine is $$W_{\text { eng }}=m c\left(T_{h}^{1 / 2}-T_{c}^{1 / 2}\right)^{2}$$

A firebox is at 750 K, and the ambient temperature is 300 K. The efficiency of a Carnot engine doing 150 J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150 J/0.600 ! 250 J from the hot reservoir and must put out 100 J of energy by heat into the environment. To follow Carnot鈥檚 reasoning, suppose that some other heat engine S could have efficiency 70.0%. (a) Find the energy input and wasted energy output of engine S as it does 150 J of work. (b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated. (c) Find the energy input and work output of engine S as it puts out exhaust energy of 100 J. (d) Let engine S operate as in (c) and contribute 150 J of its work output to running the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, the total work output, and the total energy transferred to the environment. Show that the Kelvin鈥揚lanck statement of the second law is violated. Thus our assumption about the efficiency of engine S must be false. (e) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe. Show that the entropy statement of the second law is violated.

An electric power plant that would make use of the temperature gradient in the ocean has been proposed. The system is to operate between \(20.0^{\circ} \mathrm{C}\) (surface water temperature) and \(5.00^{\circ} \mathrm{C}\) (water temperature at a depth of about 1 \(\mathrm{km}\) ). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is 75.0 MW, how much energy is taken in from the warm reservoir per hour? (c) In view of your answer to part (a), do you think such a system is worthwhile? Note that the 鈥渇uel鈥� is free.
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