Question

A closed system of mass \(5 \mathrm{~kg}\) undergoes a process in which there is work of magnitude \(9 \mathrm{~kJ}\) to the system from the surroundings. The elevation of the system increases by \(700 \mathrm{~m}\) during the process. The specific internal energy of the system decreases by \(6 \mathrm{~kJ} / \mathrm{kg}\) and there is no change in kinetic energy of the system. The acceleration of gravity is constant at \(g=9.6\) \(\mathrm{m} / \mathrm{s}^{2}\). Determine the heat transfer, in \(\mathrm{kJ}\).

Short answer

The heat transfer is \( -3.6 \text{ kJ} \).

Step-by-step solution

Identify known values

List all the given data.- Mass of the system, \( m = 5 \text{ kg} \)- Work done on the system, \( W = 9 \text{ kJ} \)- Change in elevation, \( \text{Δ}z = 700 \text{ m} \)- Change in specific internal energy, \( \text{Δ}u = -6 \text{ kJ/kg} \)- Acceleration due to gravity, \( g = 9.6 \text{ m/s}^2 \)

Calculate the change in internal energy

The change in internal energy per unit mass is given by \( \text{Δ}u = -6 \text{ kJ/kg} \). Multiply by the mass to find the total change in internal energy.\( \text{Δ}U = m \times \text{Δ}u = 5 \text{ kg} \times (-6 \text{ kJ/kg}) = -30 \text{ kJ} \)

Calculate the change in potential energy

Use the formula for potential energy change: \( \text{Δ}PE = m \times g \times \text{Δ}z \). Plug in the known values.\( \text{Δ}PE = 5 \text{ kg} \times 9.6 \text{ m/s}^2 \times 700 \text{ m} = 33600 \text{ J} \). Convert this to kJ: \( 33600 \text{ J} = 33.6 \text{ kJ} \)

Apply the First Law of Thermodynamics

Use the First Law of Thermodynamics which states \( \text{Δ}U = Q - W + \text{Δ}KE + \text{Δ}PE \). Given \( \text{Δ}KE = 0 \), solve for heat transfer, \( Q \).\( -30 \text{ kJ} = Q - 9 \text{ kJ} + 0 \text{ kJ} + 33.6 \text{ kJ} \)Rearrange the equation to solve for \( Q \).\( Q = -30 \text{ kJ} + 9 \text{ kJ} - 33.6 \text{ kJ} = -3.6 \text{ kJ} \)

Key concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed in an isolated system. Instead, the total energy of the system is conserved. The equation representing this law is: \[\text{Δ}U = Q - W + \text{Δ}KE + \text{Δ}PE\]Where: \(\text{Δ}U\) is the change in internal energy, \(Q\) is the heat added to the system, \(W\) is the work done by the system, \(\text{Δ}KE\) is the change in kinetic energy, and \( \text{Δ}PE\) is the change in potential energy.The essence of this law is that the energy entering a system must either increase the internal energy, change the kinetic or potential energy, or be used to do work.

Internal Energy

Internal Energy (\text{U}\text{)} refers to the energy contained within a system due to the kinetic and potential energies of its molecules. During a process, changes in internal energy (\text{ΔU}) occur due to heat transfer and work interactions. In our exercise, the specific internal energy of the system decreases by \(-6 \text{ kJ/kg}\). With a mass of \(5 \text{ kg}\), the total change in internal energy is:\[\text{Δ}U = m \times \text{Δ}u = 5 \text{ kg} \times (-6 \text{ kJ/kg}) = -30 \text{ kJ}\]This decrease in internal energy indicates that the system has lost some of its internal energy during the process.

Potential Energy

Potential Energy (\text{PE}\text{)} is the energy possessed by a system due to its position in a gravitational field. For an object at a height \(z\) with mass \(m\), the potential energy is given by \(PE = m \times g \times z \), where \(\text{g}\) is the acceleration due to gravity. In our problem, the change in elevation is \(\text{Δ}z = 700 \text{ m}\). The initial and final potential energies differ by:\[\text{Δ}PE = m \times g \times \text{Δ}z = 5 \text{ kg} \times 9.6 \text{ m/s}^2 \times 700 \text{ m} = 33600 \text{ J} = 33.6 \text{ kJ} \]This positive change in potential energy means the system's elevation increased, adding energy to the system in the form of gravitational potential energy.

Work-Energy Principle

The Work-Energy Principle states that the work done on or by a system results in a change in its energy. Mathematically, it can be written as: \[W = \text{Δ}E \], where \(\text{Δ}E\) is the change in the system's energy. In our exercise, the work done on the system is \(\text{W} = 9 \text{kJ}\). The heat transfer \(Q\) can be found using the First Law of Thermodynamics: \[ \text{Δ}U = Q - W + \text{Δ}PE \]Substituting the known values: \[ -30 \text{kJ} = Q - 9 \text{kJ} + 33.6 \text{kJ} \]Rearranging to solve for \(Q\), the heat transfer is: \[Q = -30 \text{kJ} + 9 \text{kJ} + 33.6 \text{kJ} = 12.6 \text{kJ} \]Thus, a heat transfer of \(12.6 \text{kJ}\) occurs during the process to balance the changes in internal energy, work done, and potential energy change.