Question

Two constant-pressure devices, each filled with \(30 \mathrm{kg}\) of air, have temperatures of \(900 \mathrm{K}\) and \(300 \mathrm{K}\). A heat engine placed between the two devices extracts heat from the high-temperature device, produces work, and rejects heat to the low-temperature device. Determine the maximum work that can be produced by the heat engine and the final temperatures of the devices. Assume constant specific heats at room temperature.

Short answer

Given the mass of air, \(m = 30 \mathrm{kg}\), and the initial temperatures, \(T_1 = 900 \mathrm{K}\) and \(T_2 = 300 \mathrm{K}\), determine the maximum work produced by the heat engine and the final temperatures of the two constant-pressure devices. The maximum work produced by the heat engine can be found using the formula: \(W_\text{max} = (U_1 - U_\text{final})(1 - \frac{T_\text{low}}{T_\text{high}})\) The final temperatures of the devices can be calculated using the relationship: \(T_\text{final} = \frac{U_\text{final}}{m \times c_v}\) By following the described steps and applying energy conservation, we can determine the maximum work and final temperatures for both devices.

Step-by-step solution

Calculate the initial internal energies of the devices

Let's first calculate the initial internal energies of the high temperature device (device 1) and the low temperature device (device 2) using the formula: \(U = m \times c_v \times T\) Where \(U\) is the internal energy, \(m\) is the mass of air, \(c_v\) is the specific heat capacity at constant volume, and \(T\) is the temperature. Given the mass, \(m = 30 \mathrm{kg}\), and the initial temperatures, \(T_1 = 900 \mathrm{K}\) and \(T_2 = 300 \mathrm{K}\). Also, since we are assuming constant specific heats at room temperature, we can use the value of \(c_v = 718 \mathrm{J/kg\cdot K}\) for air. Now, calculate the initial internal energies: \(U_1 = m \times c_v \times T_1\) \(U_2 = m \times c_v \times T_2\)

Calculate the maximum work produced by the heat engine, using the Second Law of Thermodynamics

According to the Second Law of Thermodynamics, the maximum possible work done by an engine is given by: \(W_\text{max} = Q_\text{absorbed}(1 - \frac{T_\text{low}}{T_\text{high}})\) However, for this particular case, we will need to first find the heat absorbed, which can be expressed as: \(Q_\text{absorbed} = U_1 - U_\text{final}\) \(W_\text{max}\) can then be found as: \(W_\text{max} = (U_1 - U_\text{final})(1 - \frac{T_\text{low}}{T_\text{high}})\)

Apply energy conservation to find the final internal energy of device 1

Since the engine is placed between the devices, the total energy of the system (device 1, device 2, and the heat engine) must be conserved. We know that \(U_\text{total} = U_1 + U_2\) Since energy is conserved, the final energy of the system is \(U_\text{total} = U_\text{final1} + U_\text{final2}\) In terms of heat transfer and work, the total energy can also be written as: \(Q_\text{absorbed} + Q_\text{rejected} + W_\text{max} = U_\text{final1} + U_\text{final2}\) We know that \(Q_\text{absorbed} = U_1 - U_\text{final1}\), and \(Q_\text{rejected} = U_\text{final2} - U_2\). Substituting these into the energy conservation equation, we get: \((U_1 - U_\text{final1}) + (U_\text{final2} - U_2) + W_\text{max} = U_\text{final1} + U_\text{final2}\) Solving for \(U_\text{final1}\), we find: \(U_\text{final1} = U_1 - U_2 - W_\text{max}\)

Calculate the final temperatures of the devices

With the final internal energy values in hand, we can now calculate the final temperatures for each device using the relationship: \(T_\text{final} = \frac{U_\text{final}}{m \times c_v}\) Calculate the final temperatures for device 1 and device 2: \(T_\text{final1} = \frac{U_\text{final1}}{m \times c_v}\) \(T_\text{final2} = \frac{U_\text{final2}}{m \times c_v}\) Finally, we have our result for the maximum work done by the heat engine and the final temperatures of the devices.