Question

Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, \(30 \mathrm{kJ}\) of heat is transferred to the water, and \(5 \mathrm{kJ}\) of heat is lost to the surrounding air. The paddle-wheel work amounts to \(500 \mathrm{N} \cdot \mathrm{m}\). Determine the final energy of the system if its initial energy is \(10 \mathrm{kJ}\)

Short answer

Answer: The final energy of the closed pan system is 34.5 kJ.

Step-by-step solution

Identify the known values

In this problem, we know the following values: - Heat transferred to the water: \(Q_{in} = 30 \mathrm{kJ}\) - Heat lost to the surrounding air: \(Q_{out} = 5 \mathrm{kJ}\) - Paddle-wheel work done: \(W_{paddle} = 500 \mathrm{N \cdot m}\) - Initial energy of the system: \(E_{initial} = 10 \mathrm{kJ}\)

Convert the work done to the same unit as energy

The work done (\(W_{paddle}\)) is given in \(\mathrm{N \cdot m}\), and we need to convert this to the same unit as energy, which is \(\mathrm{kJ}\). We can use the conversion factor: \(1\ \mathrm{N \cdot m} = 1\ \mathrm{J}\) and \(1 \mathrm{kJ} = 1\times10^{3} \mathrm{J}\). First, convert \(500 \mathrm{N \cdot m}\) to \(\mathrm{J}\): \(W_{paddle} = 500 \times 1 = 500\ \mathrm{J}\) Now, convert \(500 \mathrm{J}\) to \(\mathrm{kJ}\): \(W_{paddle} = \dfrac{500}{1\times 10^{3}} = 0.5 \mathrm{kJ}\)

Use the first law of thermodynamics to find the final energy

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system and the heat lost to the surroundings: \(\Delta E = Q_{in} - Q_{out} - W_{paddle}\) Using the values from steps 1 and 2, we can calculate the change in internal energy: \(\Delta E = 30 \mathrm{kJ} - 5 \mathrm{kJ} - 0.5 \mathrm{kJ} = 24.5 \mathrm{kJ}\)

Calculate the final energy of the system

Now that we have the change in internal energy, we can find the final energy of the system by simply adding the initial energy to the change in internal energy: \(E_{final} = E_{initial} + \Delta E\) \(E_{final} = 10 \mathrm{kJ} + 24.5 \mathrm{kJ} = 34.5 \mathrm{kJ}\) So, the final energy of the system is \(34.5 \mathrm{kJ}\).