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A heat engine absorbs \(52.4 \mathrm{~kJ}\) of heat and exhausts \(36.2 \mathrm{~kJ}\) of heat each cycle. Calculate \((a)\) the efficiency and \((b)\) the work done by the engine per cycle.

Short Answer

Expert verified
The efficiency of the heat engine is approximately \(0.3085\) or \(30.85 \%\), and the work done per cycle is \(16.2 \, kJ\).

Step by step solution

01

Calculate the Efficiency

First, calculate the efficiency of the heat engine. This is done using the formula \(η = 1 - \frac{Q_{out}}{Q_{in}}\), where \(Q_{in} = 52.4 \, kJ\) is the heat absorbed and \(Q_{out} = 36.2 \, kJ\) is the heat exhausted. Substitute the given values to obtain the efficiency.
02

Calculate the Work Done

Next, calculate the work done by the heat engine per cycle. The formula is \(Work \, done = Q_{in} - Q_{out}\), where \(Q_{in} = 52.4 \, kJ\) and \(Q_{out} = 36.2 \, kJ\). Substitute the given values into the equation and calculate the work done.

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Key Concepts

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

Efficiency of Heat Engines
Understanding the efficiency of heat engines is essential in thermodynamics. Efficiency is a measure of how well a heat engine converts absorbed heat into useful work. The formula for efficiency \( \eta \) is given by:\[\eta = 1 - \frac{Q_{out}}{Q_{in}}\]- \(Q_{in}\) represents the heat absorbed by the engine, in this case, 52.4 kJ.
- \(Q_{out}\) stands for the heat that is not converted into work and is exhausted, here it is 36.2 kJ.
Plugging these values into the formula allows us to find the efficiency. This efficiency value is expressed as a percentage, showing what portion of the absorbed heat is turned into work. Remember, no heat engine can be 100% efficient due to inevitable losses to friction and other factors.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. Heat engines are a practical application of thermodynamics, converting heat into mechanical work.
Thermodynamics operates under four fundamental laws, but the first and second laws are particularly relevant to understanding heat engines.
  • The First Law of Thermodynamics focuses on energy conservation, stating that energy cannot be created or destroyed, only converted.
  • The Second Law introduces the concept of entropy, explaining why energy conversion processes are never perfectly efficient.
Understanding these principles helps explain why not all absorbed heat is converted into work in heat engines, aligning with the calculated efficiency.
Work Done by Heat Engines
The work done by a heat engine is the amount of energy converted from heat into mechanical work in each cycle. To find the work done, we use the relation:\[Work \ done = Q_{in} - Q_{out}\]Here, the absorbed heat \(Q_{in}\) is 52.4 kJ, and the exhausted heat \(Q_{out}\) is 36.2 kJ. By substituting these values into the equation, we can calculate the work done, which provides valuable insight into the engine's performance.
This equation highlights a fundamental aspect of heat engines: not all absorbed heat is usable; the difference between \(Q_{in}\) and \(Q_{out}\) gives us the exact amount of energy performing useful work. This understanding is crucial in designs and improvements of energy-efficient engines.

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Most popular questions from this chapter

(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?

(a) In a two-stage Carnot heat engine, a quantity of heat \(\left|Q_{1}\right|\) is absorbed at a temperature \(T_{1}\), work \(\left|W_{1}\right|\) is done, and a quantity of heat \(\left|Q_{2}\right|\) is expelled at a lower temperature \(T_{2}\), by the first stage. The second stage absorbs the heat expelled by the first, does work \(\left|W_{2}\right|\), and expels a quantity of heat \(\left|Q_{3}\right|\) at a lower temperature \(T_{3}\). Prove that the efficiency of the combination is \(\left(T_{1}-T_{3}\right) / T_{1} .(b)\) A combination mercury-steam turbine takes saturated mercury vapor from a boiler at \(469^{\circ} \mathrm{C}\) and exhausts it to heat a steam boiler at \(238^{\circ} \mathrm{C}\). The steam turbine receives steam at this temperature and exhausts it to a condenser at \(37.8^{\circ} \mathrm{C}\). Calculate the maximum efficiency of the combination.

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.

In a Carnot cycle, the isothermal expansion of an ideal gas takes place at \(412 \mathrm{~K}\) and the isothermal compression at \(297 \mathrm{~K}\). During the expansion, \(2090 \mathrm{~J}\) of heat energy are transferred to the gas. Determine \((a)\) the work performed by the gas during the isothermal expansion, ( \(b\) ) the heat rejected from the gas during the isothermal compression, and \((c)\) the work done on the gas during the isothermal compression.

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 \((a) ?\)

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