Problem 9
Question
In a combustion cylinder, the total internal energy change produced from the burning of a fuel is \(-2573 \mathrm{~kJ}\). The cooling system that surrounds the cylinder absorbs \(947 \mathrm{~kJ}\) as heat. How much work can be done by the fuel in the cylinder during this process?
Step-by-Step Solution
Verified Answer
The work that can be done by the fuel in the cylinder during this process is 3520 kJ.
1Step 1: Understand the first law of thermodynamics
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system on the surroundings: ΔU = Q - W. This relationship will help calculate the work done.
2Step 2: Identify the given variables
From the problem, we have identified the change in internal energy of the system (ΔU) as -2573 kJ and the heat absorbed by the cooling system (Q) as 947 kJ. Here, Q is positive since heat is absorbed by the system.
3Step 3: Set up the equation for work done
Using the first law equation, rearrange to solve for work (W): W = Q - ΔU.
4Step 4: Calculate the work done
Substitute the given values into the rearranged first law equation: W = 947 kJ - (-2573 kJ). Note that subtracting a negative is the same as adding a positive.
5Step 5: Perform the calculation
Carrying out the subtraction we get: W = 947 kJ + 2573 kJ = 3520 kJ of work done by the fuel. Since we are dealing with energy, the work done is also in kilojoules.
Key Concepts
Internal Energy ChangeHeat AbsorptionWork Done Calculation
Internal Energy Change
The 'internal energy change' in a system, often symbolized as \(\text{\text\bigtriangleup}U\), refers to the difference in the total energy contained within that system at two different times. In thermodynamics, this concept is crucial as it embodies the sum of the potential and kinetic energy of the particles that make up the system. For example, when fuel is burned in a combustion cylinder, the internal energy of the system changes due to the energy released from the fuel.
The internal energy can alter through various processes, and these changes are primarily the result of heat transfer and work done on or by the system. If the energy of the system increases due to heat absorption or work done on it, the internal energy change is positive. Conversely, if the system loses energy due to heat release or work done by the system, this change is negative. In the given exercise, the negative internal energy change of \(-2573 \text{ kJ}\) indicates that the overall energy of the system has decreased due to the work done by the system on its surroundings.
The internal energy can alter through various processes, and these changes are primarily the result of heat transfer and work done on or by the system. If the energy of the system increases due to heat absorption or work done on it, the internal energy change is positive. Conversely, if the system loses energy due to heat release or work done by the system, this change is negative. In the given exercise, the negative internal energy change of \(-2573 \text{ kJ}\) indicates that the overall energy of the system has decreased due to the work done by the system on its surroundings.
Heat Absorption
The term 'heat absorption' relates to the process by which a system takes in heat energy from its surroundings. It's an essential aspect of the first law of thermodynamics, signifying the energy transfer that occurs as a result of a temperature difference. Heat can be transferred by various means, including conduction, convection, or radiation.
In the context of our problem, the cooling system around the combustion cylinder absorbs \(947 \text{ kJ}\) of heat. This is a critical detail because heat absorbed by the system contributes to the internal energy change and thus impacts the work potential of the system. It's important to understand that when a system absorbs heat, this is considered as energy being added to that system, resulting in a positive sign for the value of \(Q\). This is why, while solving the problem, heat absorbed is given a positive value, indicating an increase in the system's energy – contrary to the internal energy change which is negative in this case.
In the context of our problem, the cooling system around the combustion cylinder absorbs \(947 \text{ kJ}\) of heat. This is a critical detail because heat absorbed by the system contributes to the internal energy change and thus impacts the work potential of the system. It's important to understand that when a system absorbs heat, this is considered as energy being added to that system, resulting in a positive sign for the value of \(Q\). This is why, while solving the problem, heat absorbed is given a positive value, indicating an increase in the system's energy – contrary to the internal energy change which is negative in this case.
Work Done Calculation
Calculating the 'work done' is a fundamental application of the first law of thermodynamics. It can be visualized as the energy required to move an object or produce a change within a system. The ability to calculate this work is important in understanding how much energy systems can exert on their surroundings.
To compute work done in thermodynamic processes, we use the formula \(W = Q - \text{\text\bigtriangleup}U\), where \(W\) is the work done, \(Q\) is the heat added to the system, and \(\text{\text\bigtriangleup}U\) is the change in internal energy. From the exercise, rearranging the first law of thermodynamics allows us to find that the work done by the fuel in the combustion cylinder is the sum of the heat absorbed by the cooling system and the negative of the internal energy change. By plugging in the given values, we find that the fuel can do \(3520 \text{ kJ}\) of work, which reflects the total energy available for doing work after accounting for the heat absorbed in the combustion process.
Understandably, heat and work are interrelated; work done can result in heat transfer and vice versa. This interplay is central to energy transformation processes in thermal systems and engineering applications.
To compute work done in thermodynamic processes, we use the formula \(W = Q - \text{\text\bigtriangleup}U\), where \(W\) is the work done, \(Q\) is the heat added to the system, and \(\text{\text\bigtriangleup}U\) is the change in internal energy. From the exercise, rearranging the first law of thermodynamics allows us to find that the work done by the fuel in the combustion cylinder is the sum of the heat absorbed by the cooling system and the negative of the internal energy change. By plugging in the given values, we find that the fuel can do \(3520 \text{ kJ}\) of work, which reflects the total energy available for doing work after accounting for the heat absorbed in the combustion process.
Understandably, heat and work are interrelated; work done can result in heat transfer and vice versa. This interplay is central to energy transformation processes in thermal systems and engineering applications.
Other exercises in this chapter
Problem 6
(a) Calculate the work for a system that releases \(346 \mathrm{~kJ}\) of heat in a process for which the decrease in internal energy is \(125 \mathrm{~kJ}\). (
View solution Problem 7
A gas in a cylinder was placed in a heater and gained \(7000 \mathrm{~kJ}\) of heat. If the cylinder increased in volume from \(700 \mathrm{~mL}\) to \(1450 \ma
View solution Problem 11
For a certain reaction at constant pressure, \(\Delta H=-15 \mathrm{~kJ}\) and \(22 \mathrm{~kJ}\) of expansion work is done on the system. What is \(\Delta U\)
View solution Problem 12
For a certain reaction at constant pressure, \(\Delta U=-95 \mathrm{~kJ}\) and \(56 \mathrm{~kJ}\) of expansion work is done by the system. What is \(\Delta H\)
View solution