Question

A closed system of mass \(20 \mathrm{~kg}\) undergoes a process in which there is a heat transfer of \(1000 \mathrm{~kJ}\) from the system to the surroundings. The work done on the system is \(200 \mathrm{~kJ}\). If the initial specific internal energy of the system is \(300 \mathrm{~kJ} / \mathrm{kg}\), what is the final specific internal energy, in \(\mathrm{kJ} / \mathrm{kg}\) ? Neglect changes in kinetic and potential energy.

Short answer

The final specific internal energy is 260 \( \mathrm{kJ} / \mathrm{kg} \).

Step-by-step solution

State the First Law of Thermodynamics

The first law of thermodynamics can be expressed for a closed system as: \[ \Delta U = Q - W \] where \( \Delta U \) is the change in internal energy, \( Q \) is the heat transfer, and \( W \) is the work done by the system. In this problem, heat is transferred from the system to the surroundings, so \( Q \) will be negative.

Calculate the Change in Internal Energy

Given:- Mass \( m = 20 \mathrm{~kg} \)- Heat transfer \( Q = -1000 \mathrm{~kJ} \) (since it is transferred from the system)- Work done on the system \( W = -200 \mathrm{~kJ} \) (since work is done on the system, it is negative) Using the formula: \[ \Delta U = Q - W \]Substitute the given values:\[ \Delta U = -1000 \mathrm{~kJ} - (-200 \mathrm{~kJ}) = -1000 \mathrm{~kJ} + 200 \mathrm{~kJ} = -800 \mathrm{~kJ} \]

Calculate the Initial Internal Energy

The initial specific internal energy is given as \( u_i = 300 \mathrm{~kJ} / \mathrm{kg} \). To find the initial internal energy \( U_i \), multiply the specific internal energy by the mass of the system:\[ U_i = u_i \times m = 300 \mathrm{~kJ} / \mathrm{kg} \times 20 \mathrm{~kg} = 6000 \mathrm{~kJ} \]

Calculate the Final Internal Energy

The change in internal energy \( \Delta U \) is the difference between the final and initial internal energies. If \( U_f \) is the final internal energy, then:\[ \Delta U = U_f - U_i \]Solving for \( U_f \):\[ U_f = \Delta U + U_i = -800 \mathrm{~kJ} + 6000 \mathrm{~kJ} = 5200 \mathrm{~kJ} \]

Calculate the Final Specific Internal Energy

To find the final specific internal energy \( u_f \), divide the final internal energy by the mass of the system:\[ u_f = \frac{U_f}{m} = \frac{5200 \mathrm{~kJ}}{20 \mathrm{~kg}} = 260 \mathrm{~kJ} / \mathrm{kg} \]

Key concepts

Closed System

In thermodynamics, a closed system is a physical system that does not exchange any matter with its surroundings. However, it can exchange energy in the form of heat or work.

The given exercise involves a closed system with a mass of 20 kg. Since it is a closed system, the mass remains constant throughout the process. It is crucial to note that, unlike an open system, no mass crosses the system boundary during the process.

Understanding these properties of a closed system helps in accurately applying the First Law of Thermodynamics to solve for internal energy changes.

Heat Transfer

Heat transfer is the movement of thermal energy from one place to another. In the context of the given exercise, there is a heat transfer of 1000 kJ from the system to its surroundings.

In thermodynamics, heat transfer to the surroundings is usually considered negative, as the system loses energy. Hence, the value of Q is taken as -1000 kJ.

When solving the problem, understanding the direction of heat transfer is essential. It affects the change in internal energy of the system according to the First Law of Thermodynamics.

Internal Energy

Internal energy is the total energy contained within a system. It includes all the kinetic and potential energies of the molecules inside the system.

In the exercise, the initial specific internal energy is given as 300 kJ/kg. To find the total internal energy, multiply this value by the mass of the system: \[ U_i = u_i \times m = 300 \text{ kJ/kg} \times 20 \text{ kg} = 6000 \text{ kJ} \] The change in internal energy is calculated using \[ \Delta U = Q - W \] Given Q = -1000 kJ and W = -200 kJ, \[ \Delta U = -1000 \text{ kJ} - (-200 \text{ kJ}) = -800 \text{ kJ} \] Finally, the final internal energy can be determined by: \[ U_f = U_i + \Delta U = 6000 \text{ kJ} + (-800 \text{ kJ}) = 5200 \text{ kJ} \] To get the final specific internal energy: \[ u_f = \frac{U_f}{m} = \frac{5200 \text{ kJ}}{20 \text{ kg}} = 260 \text{ kJ/kg} \] Understanding internal energy helps us apply the First Law of Thermodynamics effectively and find the state of the system after any process.