Work Done Calculation on Liquid System

This problem involves calculating the work done on a liquid that is stirred in a well-insulated chamber, leading to a rise in its temperature. We will use the principles of thermodynamics, specifically the First Law of Thermodynamics, to determine the amount of work.

Understanding the Thermodynamics Concepts

To accurately solve this problem, it is important to understand a few core thermodynamic concepts:

• Well-insulated Chamber: A well-insulated chamber implies that there is no heat transfer between the liquid (system) and its surroundings. In thermodynamics, such a process is known as an adiabatic process, meaning the heat transfer (\(Q\)) is zero.
• First Law of Thermodynamics: This fundamental law states that the change in the internal energy (\(\Delta U\)) of a system is equal to the heat added to the system (\(Q\)) minus the work done by the system (\(W\)). Mathematically, it is expressed as: \[\Delta U = Q - W\] However, if we define work done on the system as \(W_{on}\), the equation becomes: \[\Delta U = Q + W_{on}\]
• Internal Energy Change of a Liquid: For an incompressible substance like a liquid, the change in internal energy due to a temperature change can be calculated using its mass, specific heat, and temperature difference. The formula is: \[\Delta U = m \cdot c \cdot \Delta T\] where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature.

Given Parameters for the Liquid System

Let's list the known values provided in the question for the liquid:

Note that a temperature change of 15°C is equivalent to a temperature change of 15 K, as the size of a degree Celsius is the same as the size of a Kelvin. Therefore, \(\Delta T = 15 \text{ K}\).

Applying the First Law and Calculating Work Done

Since the chamber is well-insulated, the heat transfer (\(Q\)) is zero. This simplifies the First Law of Thermodynamics significantly.

1. Calculate the Change in Internal Energy (\(\Delta U\))

Using the formula for the change in internal energy of the liquid: \[\Delta U = m \cdot c \cdot \Delta T\] Substitute the given values: \[\Delta U = (2 \text{ kg}) \cdot (3 \text{ kJ/kg-K}) \cdot (15 \text{ K})\] \[\Delta U = 6 \text{ kJ/K} \cdot 15 \text{ K}\] \[\Delta U = 90 \text{ kJ}\] This means the internal energy of the liquid increased by 90 kJ.

2. Determine the Work Done on the System (\(W_{on}\))

Now, apply the First Law of Thermodynamics, considering that work is done on the system: \[\Delta U = Q + W_{on}\] Since the chamber is well-insulated, \(Q = 0\): \[\Delta U = 0 + W_{on}\] \[\Delta U = W_{on}\] Substitute the calculated value of \(\Delta U\): \[90 \text{ kJ} = W_{on}\] Therefore, the work done on the liquid (system) is 90 kJ. This positive value indicates that energy in the form of work was added to the system, causing its internal energy to increase and thus its temperature to rise.

Final Work Done Result

The amount of work done on the liquid (or system) is 90 kJ. This matches the provided correct answer.