Problem 21 - University Physics with Modern Physics

Question

20.21. A Carnot heat engine has a thermal efficiency of 0.600 , and the temperature of its hot reservoir is 800 \(\mathrm{K}\) . If 3000 \(\mathrm{J}\) of heat is rejected to the cold reservoir in one cycle, what is the work output of the engine during one cycle?

Step-by-Step Solution

Answer

The work output of the engine is 4500 J.

Step 1: Understand the Carnot Efficiency

The Carnot efficiency (\(\eta_c\)) of a heat engine is given by the formula \[ \eta_c = 1 - \frac{T_c}{T_h} \]where \(T_c\) is the temperature of the cold reservoir, and \(T_h\) is the temperature of the hot reservoir. We are given the Carnot efficiency (\(\eta_c = 0.600\)) and the temperature of the hot reservoir (\(T_h = 800 \, \text{K}\)).

Step 2: Set Up the Efficiency Equation

Using the Carnot efficiency formula, we can rearrange it to find the cold reservoir temperature \(T_c\):\[ 0.600 = 1 - \frac{T_c}{800} \]

Step 3: Solve for Temperature of Cold Reservoir

Rearranging the equation \(0.600 = 1 - \frac{T_c}{800}\) to solve for \(T_c\):\[\frac{T_c}{800} = 1 - 0.600 = 0.400\]\[T_c = 0.400 \times 800 = 320 \, \text{K}\]

Step 4: Relationship Between Efficiencies, Work, and Heat

The efficiency of the engine is also defined as the ratio of the work output \(W\) to the heat input \(Q_h\):\[\eta_c = \frac{W}{Q_h}\]Using the fact that the waste heat \(Q_c = 3000 \, \text{J}\), and the relationship \(Q_h = W + Q_c\), substitute to find \(W\):

Step 5: Calculate the Work Output

First find the heat input:Using \(\eta_c = \frac{W}{Q_h}\), express \(Q_h\):\[Q_h = \frac{W}{\eta_c}\]Since \(\eta_c = 0.600\), and rearranging the expression \(Q_h = W + Q_c\), we have:\[Q_h = W + 3000\]\[W = \eta_c \times (W + 3000)\]\[W = 0.600 \times (W + 3000)\]\[W = 0.600W + 1800\]Solving:\[W - 0.600W = 1800\]\[0.400W = 1800\]\[W = \frac{1800}{0.400} = 4500 \, \text{J}\]

Step 6: Final Calculation and Answer

The work output \(W\) for the engine during one cycle is therefore:\[4500 \, \text{J}\]

Key Concepts

Thermal Efficiency of a Carnot Heat Engine

The thermal efficiency (\(\eta\)) of a Carnot heat engine is a measure of how well it converts heat into useful work. It is expressed as a fraction of the total input energy from the hot reservoir that is converted into work. The formula to calculate the thermal efficiency of a Carnot engine is given by:\[\eta_c = 1 - \frac{T_c}{T_h}\]where \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot reservoir. This formula indicates that higher efficiencies are achieved when the temperature difference between the hot and cold reservoirs is greater. For the specific scenario where the efficiency is given as 0.600, it means that 60% of the heat energy supplied by the hot reservoir is converted to work on each cycle.
In practical terms, thermal efficiency helps to evaluate how effectively an engine uses the heat supplied to it, guiding improvements in engine design and operation.

Understanding the Hot Reservoir

The hot reservoir in a Carnot engine is where heat is absorbed at a constant high temperature. It serves as the starting point for the engine's energy cycle. The temperature of the hot reservoir, denoted as \(T_h\), is crucial for determining the efficiency of the engine. In this problem, the hot reservoir has a temperature of 800 K.

• A higher \(T_h\) generally leads to better engine efficiency according to the Carnot efficiency equation.
• Heat energy from this reservoir is used to do useful work and is essential for maintaining the cyclic operation of the Carnot engine.

In essence, the hot reservoir is the source of energy for the Carnot cycle and its temperature directly impacts the performance of the engine. The greater the temperature, the greater the potential for efficiency when converting heat into work.

Role of the Cold Reservoir

The cold reservoir is the component of the Carnot heat engine that absorbs the expelled heat energy. It operates at a colder temperature than the hot reservoir, denoted as \(T_c\). In our case, after solving, \(T_c\) was found to be 320 K.

• This reservoir plays an integral role in determining the overall efficiency of the engine.
• The smaller the temperature difference between \(T_c\) and \(T_h\), the lower the efficiency.

The cold reservoir prevents the system from overheating by rejecting unused heat, making it as important as the hot reservoir in ensuring consistent operation. Proper understanding of \(T_c\) helps in assessing an engine’s performance and designing it for optimal functionality.

Calculating the Work Output

Work output (\(W\)) is the useful energy produced by the Carnot heat engine during one cycle of operation. It is the energy that is converted from heat and can be used to do tasks, such as turning a wheel or running a generator. To find the work output, we use the efficiency formula:\[\eta_c = \frac{W}{Q_h}\]where \(Q_h\) is the heat input. Rearranging the terms, we solve for work \(W\):\[W = \eta_c \times Q_h\]Given the problem, the efficiency is 0.600, and after calculations, the work output amounts to 4500 J. This represents the fraction of input heat energy that is converted to work, emphasizing the efficiency of the Carnot cycle in utilizing the available energy.

Heat Transfer in the Carnot Cycle

Heat transfer is a critical process in the operation of a Carnot heat engine. It involves the movement of thermal energy from one reservoir to another, and ultimately to the engine's work output. There are several key elements to understand about heat transfer in this context:

• The hot reservoir transfers heat \(Q_h\) to the engine, which is part of the energy conversion cycle.
• The engine converts a part of this heat into work, denoted as \(W\).
• The leftover heat \(Q_c\) is expelled to the cold reservoir as waste heat.

In this exercise, \(Q_c\) was given as 3000 J. Knowing the values of \(Q_h\) and the efficiency, we determined that the heat input must be 7500 J. Understanding how heat is transferred and utilized within the heat engine aids in optimizing thermal processes for improved energy efficiency.