Question

With each cycle, a 2500.-W engine extracts 2100. J from a thermal reservoir at \(90.0^{\circ} \mathrm{C}\) and expels \(1500 .\) J into a thermal reservoir at \(20.0^{\circ} \mathrm{C}\). What is the work done for each cycle? What is the engine's efficiency? How much time does each cycle take?

Short answer

Answer: In one cycle, the engine does 600 J of work, has an efficiency of 28.57%, and the cycle takes 0.24 seconds.

Step-by-step solution

Determine the work done during each cycle

Since this is a heat engine, we can use the first law of thermodynamics to determine the work done during each cycle: $$W = Q_h - Q_c$$ where \(W\) is the work done, \(Q_h\) is the heat extracted from the hot reservoir, \(Q_c\) is the heat expelled into the cold reservoir. In this case, \(Q_h = 2100\) J and \(Q_c = 1500\) J, so we have: $$W = 2100 \,\text{J} - 1500 \, \text{J} = 600 \,\text{J}$$ The work done during each cycle is 600 J.

Calculate the engine's efficiency

The efficiency of a heat engine is defined as the ratio of the work done to the heat extracted from the hot reservoir: $$\text{Efficiency} = \frac{W}{Q_h}$$ Using the values we found in Step 1, we can calculate the engine's efficiency: $$\text{Efficiency} = \frac{600 \, \text{J}}{2100\, \text{J}} = \frac{2}{7}$$ Converting this to a percentage, we have: $$\text{Efficiency} = \frac{2}{7} \times 100\% = 28.57\%$$ The engine's efficiency is 28.57%.

Determine the time taken for each cycle

To find the time taken for each cycle, we can use the given power of the engine and the work done during each cycle: $$P = \frac{W}{t}$$ where \(P\) is the power (2500 W), \(W\) is the work done during each cycle (600 J), \(t\) is the time taken for each cycle. Rearranging this formula to solve for \(t\), we get: $$t = \frac{W}{P}$$ Now, plug in the given values: $$t = \frac{600 \, \text{J}}{2500 \,\text{W}} = 0.24 \,\text{s}$$ The time taken for each cycle is 0.24 seconds.

Key concepts

Heat Engine

A heat engine is a fascinating device that operates by converting heat energy into mechanical work. The basic principle behind a heat engine is that it takes in energy from a high-temperature source, performs some work like turning a turbine, and then releases some leftover energy into a cooler environment. This process involves a continual cycle that can be defined by its input and output of heat, as well as the work it does during each cycle.

In the context of our exercise, the heat engine extracts 2100 J of energy from a hot reservoir and expels 1500 J into a cooler reservoir. The difference between the input and output heat energy is the mechanical work done, which is calculated using the first law of thermodynamics. Each cycle of this engine turns 600 J of energy into work. By constantly going through the cycle of drawing in heat, doing work, and expelling waste heat, the engine performs its function of turning thermal energy into useful work.

• Extracts heat energy from a reservoir.
• Performs work by converting some extracted energy.
• Expels unused energy into a cooler reservoir.

Efficiency Calculation

Efficiency in the context of a heat engine is a measure of how well the engine converts heat energy into mechanical work. It is the ratio of the useful work output to the heat input from the hot reservoir, usually expressed as a percentage. In many real-world applications, having a higher efficiency means less waste and better performance.

The efficiency can be calculated with the formula:\[\text{Efficiency} = \frac{W}{Q_h}\]where:

• \(W\) is the work done, which is the energy used to accomplish a task during each cycle. In this example, \(W = 600 \, \text{J}\).
• \(Q_h\) is the heat extracted from the hot reservoir, which is \(2100 \, \text{J}\) in this scenario.

Using our specific example, the efficiency calculation looks like:\[\text{Efficiency} = \frac{600 \, \text{J}}{2100 \, \text{J}} = \frac{2}{7}\]When this fraction is converted to a percentage, it equals approximately 28.57%. This value tells us that the engine turns only about 28.57% of the heat it extracts into useful work. Even though the number might not seem exceedingly high, it's important to remember that a significant portion of energy in practical engines goes to waste due to inefficiencies in the system.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, is central to understanding heat engines. It tells us that the total energy in a closed system is constant. Energy can neither be created nor destroyed, only transferred or converted from one form to another.

For a heat engine, this principle is key in balancing the energy equation. As the engine performs its cycle, the energy input from the hot reservoir \(Q_h\), minus the waste heat expelled \(Q_c\), gives the mechanical work \(W\) done. Mathematically, this is stated as:\[W = Q_h - Q_c\]From our exercise:

• Heat extracted from the hot reservoir, \(Q_h = 2100 \, \text{J}\)
• Heat expelled to the cold reservoir, \(Q_c = 1500 \, \text{J}\)
• Therefore, the work done, \(W = 2100 \, \text{J} - 1500 \, \text{J} = 600 \, \text{J}\)

This calculation shows how the first law of thermodynamics keeps track of energy transitions, ensuring that all incoming energy is accounted for either as work performed or as heat expelled.