Understanding the Thermodynamics of a Closed System at Constant Volume

Let's analyze the given thermodynamics problem involving a closed system undergoing a constant volume process. We are asked to determine the work done during this process, given the heat transfer, the temperature change, and the properties of the substance within the system.

Key Information Provided:

• System Type: Closed system
• Process Type: Constant volume (isochoric)
• Temperature rise ($\Delta T$): $50^\circ\text{C}$
• Heat transferred ($Q$): $100\text{ kJ}$ (Assuming heat is added to the system, $Q = +100\text{ kJ}$)
• Specific heat at constant volume ($c_v$): $1\text{ kJ/kg }^\circ\text{C}$
• Mass of substance ($m$): $3\text{ kg}$

Applying the First Law of Thermodynamics

For a closed system, the First Law of Thermodynamics states that the change in internal energy ($\Delta U$) is equal to the heat added to the system ($Q$) minus the work done by the system ($W$). Mathematically, this is expressed as:

$\Delta U = Q - W$

We need to find $W$. To do this, we first need to calculate the change in internal energy, $\Delta U$.

Calculating the Change in Internal Energy ($\Delta U$)

The change in internal energy for a substance with constant specific heat at constant volume ($c_v$) is given by:

$\Delta U = m \cdot c_v \cdot \Delta T$

Let's substitute the given values into this equation:

• Mass ($m$) = $3\text{ kg}$
• Specific heat at constant volume ($c_v$) = $1\text{ kJ/kg }^\circ\text{C}$
• Temperature rise ($\Delta T$) = $50^\circ\text{C}$

So, the change in internal energy is:

$\Delta U = (3\text{ kg}) \cdot (1\text{ kJ/kg }^\circ\text{C}) \cdot (50^\circ\text{C})$

$\Delta U = 150\text{ kJ}$

Determining Work Done ($W$)

Now we can use the First Law of Thermodynamics ($\Delta U = Q - W$) to find the work done ($W$). We have $\Delta U = 150\text{ kJ}$ and $Q = 100\text{ kJ}$.

Rearranging the First Law equation to solve for $W$:

$W = Q - \Delta U$

Substitute the values:

$W = 100\text{ kJ} - 150\text{ kJ}$

$W = -50\text{ kJ}$

Understanding Work in a Constant Volume Process

It is important to note that for a constant volume (isochoric) process in a simple compressible system where no other forms of work (like electrical or shaft work) are mentioned, the boundary work ($W_{boundary} = \int P \cdot dV$) is typically zero because $dV=0$. However, the First Law $\Delta U = Q - W$ accounts for all forms of work. If only boundary work was present, $W$ would be zero. Since $Q$ is non-zero and $\Delta U$ is non-zero, $W$ must also be non-zero according to the First Law. The calculated value $W = -50\text{ kJ}$ indicates that there must be work done on the system (e.g., stirring work) which is not explicitly mentioned but is implied by the non-zero $Q$ and $\Delta U$ in this constant volume process scenario.

Summary of Calculations:

The work done in this case is -50 kJ.