Question

Two kilograms of air is contained in a rigid well-insulated tank with a volume of \(0.6 \mathrm{~m}^{3}\). The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of \(10 \mathrm{~W}\) for \(1 \mathrm{~h}\). If no changes in kinetic or potential energy occur, determine (a) the specific volume at the final state, in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the energy transfer by work, in \(\mathrm{kJ}\). (c) the change in specific internal energy of the air, in \(\mathrm{kJ} / \mathrm{kg}\).

Short answer

Specific volume: \(0.3 \mathrm{~m}^{3}/\mathrm{kg}\), Energy transfer by work: \(36 \mathrm{kJ}\), Change in specific internal energy: \(-18 \mathrm{kJ}/\mathrm{kg}\).

Step-by-step solution

Calculate the specific volume at initial state

Given the volume of the tank is \(0.6 \mathrm{~m}^{3}\) and the mass of the air is \(2 \mathrm{~kg}\). The specific volume is defined as volume per unit mass: \ \[v = \frac{V}{m} = \frac{0.6 \mathrm{~m}^{3}}{2 \mathrm{~kg}} = 0.3 \mathrm{~m}^{3}/\mathrm{kg}\] Thus, the specific volume at the initial state is \(0.3 \mathrm{~m}^{3}/\mathrm{kg}\).

Calculate the energy transfer by work

The power transferred by the paddle wheel is given as \(10 \mathrm{~W}\) and the duration is \(1 \mathrm{~h}\). First, convert the time into seconds: \ \[1 \mathrm{~h} = 3600 \mathrm{~s}\] \ The energy transferred by work is the product of power and time: \ \[W = P \times t = 10 \mathrm{~W} \times 3600 \mathrm{~s} = 36000 \mathrm{~J} = 36 \mathrm{kJ}\] Hence, the energy transfer by work is \(36 \mathrm{kJ}\).

Calculate the change in specific internal energy

Since the process occurs in a well-insulated tank with no changes in kinetic or potential energy, the first law of thermodynamics dictates that the energy added by work results in a change in internal energy: \ \[\Delta U = Q - W\] Since the tank is well-insulated, \(Q = 0\): \ \[\Delta U = -W = -36 \mathrm{kJ}\] The change in specific internal energy, \(\Delta u\), is given by: \ \[\Delta u = \frac{\Delta U}{m} = \frac{-36 \mathrm{kJ}}{2 \mathrm{~kg}} = -18 \mathrm{kJ}/\mathrm{kg}\] Therefore, the change in specific internal energy of the air is \(-18 \mathrm{kJ}/\mathrm{kg}\).

Key concepts

Specific Volume Calculation

Understanding specific volume is crucial in solving thermodynamics problems. Specific volume is defined as the volume occupied by a unit mass of a substance. Mathematically, it is represented as:
\( v = \frac{V}{m} \)
where V is the volume and m is the mass.

In our given exercise, we have a volume of 0.6 \( \text{m}^3 \) and a mass of 2 kg. By dividing the volume by the mass, we get:
\[ v = \frac{0.6~\text{m}^3}{2~\text{kg}} = 0.3~\text{m}^3/\text{kg} \]
Thus, the specific volume of air at the initial state is 0.3 \( \text{m}^3/\text{kg} \).

Specific volume helps us understand how dispersed or concentrated the mass is within a given volume. It plays a key role in analyzing various thermodynamics problems, particularly those involving gases and liquids.”

Energy Transfer by Work

Energy transfer by work is another fundamental concept in thermodynamics. When work is done on or by a system, energy is transferred. In our problem, a paddle wheel transfers energy to air in the tank.

The power (rate of energy transfer) provided by the paddle wheel is given as 10 W (watts). This power is sustained for a period of 1 hour. To find the total energy transferred, we first convert the time to seconds:
\[ 1 \text{ hour} = 3600 \text{ seconds} \]
Now, we multiply the power by the time to get the total energy transferred:
\[ W = P \times t = 10 \text{ W} \times 3600 \text{ s} = 36000 \text{ J} = 36 \text{ kJ} \]
Therefore, the energy transfer by work is 36 kJ.

This concept helps in understanding how systems exchange energy with their surroundings. It's essential in the study of engines, refrigerators, and various other systems where work and energy interactions are fundamental.”
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Change in Specific Internal Energy

Specific internal energy represents the energy stored within a system on a per mass basis. For our problem, no kinetic or potential energy changes take place. The system is well-insulated, meaning no heat exchange occurs. According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Here, \( Q \) is zero because the tank is insulated:
\[ \Delta U = -W \]
We've calculated the energy transfer by work to be 36 kJ. Therefore:
\[ \Delta U = -36 \text{ kJ} \]
To find the change in specific internal energy, we divide by the mass:
\[ \Delta u = \frac{ \Delta U}{m} = \frac{-36 \text{ kJ}}{2 \text{ kg}} = -18 \text{ kJ/kg} \]
Thus, the change in specific internal energy of the air is \ \ -18 \text{ kJ/kg} \ \ .

This concept provides insight into how energy within a system changes due to work being done on or by the system. It’s vital for understanding how energy transformations occur in thermodynamic processes.”
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