Question

A system undergoes a thermodynamic power cycle while receiving energy by heat transfer from an incompressible body of mass \(m\) and specific heat \(c\) initially at temperature \(T_{\mathrm{H}}\). The system undergoing the cycle discharges energy by heat transfer to another incompressible body of mass \(m\) and specific heat \(c\) initially at a lower temperature \(T_{\mathrm{C}}\). Work is developed by the cycle until the temperature of each of the two bodies is the same, \(T^{\prime}\). (a) Develop an expression for the minimum theoretical final temperature, \(T^{\prime}\), in terms of \(m, c, T_{\mathrm{H}}\), and \(T_{\mathrm{C}}\), as required. (b) Develop an expression for the maximum theoretical amount of work that can be developed, \(W_{\max }\), in terms of \(m, c, T_{\mathrm{H}}\), and \(T_{\mathrm{C}}\), as required. (c) What is the minimum theoretical work input that would be required by a refrigeration cycle to restore the two bodies from temperature \(T^{\prime}\) to their respective initial temperatures, \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}} ?\)

Short answer

a) \( T' = \frac{T_H + T_C}{2} \)b) \( W_{max} = m \times c \times (T_H - T_C) \)c) \( W_{min} = m \times c \times (T_H - T_C) \)

Step-by-step solution

Understand the Problem

Two bodies with mass \(m\) and specific heat \(c\) undergo a thermodynamic cycle until they reach a common temperature \(T'\). The bodies start at temperatures \(T_H\) and \(T_C\). The goal is to find the equilibrium temperature \(T'\), the maximum work \(W_{max}\) that can be developed, and the minimum work input required to restore initial temperatures.

Establish Energy Balance

Because the total energy of the system is conserved, equate the heat lost by the hot body to the heat gained by the cold body. Set up the equation: \[ Q_H = Q_C \] Where the heat transfer can be expressed as: \[ m \times c \times (T_H - T') = m \times c \times (T' - T_C) \]

Solve for the Final Temperature \(T'\)

Simplify the equation from Step 1 to find \(T'\): \[ m \times c \times (T_H - T') = m \times c \times (T' - T_C) \] Since \(m\) and \(c\) are the same for both bodies, the equation reduces to: \[ T_H - T' = T' - T_C \] Rearrange to solve for \(T'\): \[ 2T' = T_H + T_C \] Thus, \[ T' = \frac{T_H + T_C}{2} \]

Determine the Maximum Work \(W_{max}\)

The maximum work done is equal to the heat transfer from the hot body minus the heat transfer to the cold body when they reach equilibrium at \(T'\). Calculate the work as follows: \[ W_{max} = Q_H - Q_C = m \times c \times (T_H - T') - m \times c \times (T' - T_C) \] Substitute \(T'\) from Step 2: \[ W_{max} = m \times c \times \left(T_H - \frac{T_H + T_C}{2}\right) - m \times c \times \left(\frac{T_H + T_C}{2} - T_C\right) \] Simplify the expression: \[ W_{max} = m \times c \times \left( \frac{T_H - T_C}{2} \right) + m \times c \times \left( \frac{T_H - T_C}{2} \right) \] Combine terms to get: \[ W_{max} = m \times c \times (T_H - T_C) \]

Calculate Minimum Work Input for Refrigeration Cycle

To restore the bodies to their initial temperatures, you need to reverse the process, effectively moving heat from the colder body to the hotter one. Given the principles of thermodynamics, the minimum work input \(W_{min}\) for this refrigeration cycle is the same as the maximum work done when they were brought to equilibrium: \[ W_{min} = W_{max} = m \times c \times (T_H - T_C) \]

Key concepts

Energy Balance

In thermodynamic cycles, energy conservation is key. When two bodies exchange heat until they reach an equilibrium temperature, the energy lost by the hotter body equals the energy gained by the colder body. This principle of energy balance is expressed as:
\ Q_H = Q_C \
Here, the heat transferred can be written as:
\ m \times c \times (T_H - T') = m \times c \times (T' - T_C) \
Notice that since both bodies have the same mass \(m\) and specific heat \(c\), these terms cancel out, simplifying our equations. This showcases conservation of energy in basic thermodynamic processes, ensuring that no heat is lost to the environment.

Specific Heat

Specific heat \( c \) is a property that indicates how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree. In our problem, both bodies have the same specific heat, making calculations straightforward due to symmetry. Here's the general formula:
\ Q = m \times c \times \Delta T \
Where:

• \(Q\) is the heat added or removed
• \(m\) is the mass of the substance
• \(c\) is the specific heat capacity
• \Delta T\ is the change in temperature

Specific heat is crucial in determining how the temperature of each body changes during energy transfer. Small specific heat values mean temperature changes rapidly with heat addition or removal. Larger specific heat indicates slower temperature changes.

Equilibrium Temperature

Equilibrium is reached when both bodies involved in the heat exchange end up at the same temperature, \(T'\), meaning no further net heat transfer occurs. Practically, you can think of this as both bodies leveling out their thermal energy. To find \(T'\):
\ T' = \frac{T_H + T_C}{2} \
This average temperature formula is derived by setting the total heat lost by the hot body as equal to the total heat gained by the cold body. Since the mass and specific heat are equal, we get a simple average, representing the final resting point of their temperatures.

Maximum Work

The maximum theoretical work \(W_{max}\) that can be developed by the thermodynamic power cycle is critical. It represents the usable energy derived from the heat transfer before reaching equilibrium. Calculated as:
\ W_{max} = m \times c \times (T_H - T_C) \
This formula comes from the difference in heat content times the capacities of the bodies. It reveals that the greater the temperature difference between initial bodies, the more work you can derive from the system. This showcases the potential energy conversion during the cycle.

Refrigeration Cycle

A refrigeration cycle does the reverse work of a thermodynamic power cycle. To restore the bodies to their initial temperatures, it requires an external work input \(W_{min}\). Essentially, this cycle moves heat from the cooler body back to the hotter one. Usually, it is calculated using:
\ W_{min} = W_{max} = m \times c \times (T_H - T_C) \
This equality stems from thermodynamic reversibility, meaning the energy to reverse the process equals the energy gained moving forward. Thus, a refrigeration cycle needs the same amount of input work to restore original conditions, highlighting energy efficiency in temperature control processes.