Question

A cylindrical rod of length \(L\) insulated on its lateral surface is initially in contact at one end with a wall at temperature \(T_{\mathrm{H}}\) and at the other end with a wall at a lower temperature \(T_{\mathrm{C}}\). The temperature within the rod initially varies linearly with position \(z\) according to $$ T(z)=T_{\mathrm{H}}-\left(\frac{T_{\mathrm{H}}-T_{\mathrm{C}}}{L}\right) z $$ The rod is then insulated on its ends and eventually comes to a final equilibrium state where the temperature is \(T_{\mathrm{f}}\). Evaluate \(T_{\mathrm{f}}\) in terms of \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) and show that the amount of entropy produced is $$ \sigma=m c\left(1+\ln T_{\mathrm{f}}+\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} \ln T_{\mathrm{C}}-\frac{T_{\mathrm{H}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} \ln T_{\mathrm{H}}\right) $$ where \(c\) is the specific heat of the rod.

Short answer

The final temperature \(T_{f}\) is \frac{T_{H} + T_{C}}{2}\. The entropy produced is \ m c \left(1 + \ln T_{f} + \frac{T_{C}}{T_{H} - T_{C}} \ln T_{C} - \frac{T_{H}}{T_{H} - T_{C}} \ln T_{H} \right) \.

Step-by-step solution

Determine the Final Temperature

Since the rod comes to a final equilibrium state, the final temperature \(T_{f}\) must be the average of the initial temperatures at the two ends. Calculate \(T_{f}\) using the formula: \[ T_{f} = \frac{T_{H} + T_{C}}{2} \]

Calculate the Initial Entropy

The initial temperature distribution is linear. Use the temperature profile \[ T(z) = T_{H} - \left( \frac{T_{H}-T_{C}}{L} \right) z \] to determine the initial entropy. The initial entropy is not uniform due to the variation in temperature.

Determine the Heat Capacity

Assume the specific heat capacity \(c\) is constant. The total heat energy of the rod remains constant during the process.

Calculate Final Entropy

The final entropy \(S_{f}\) when the rod reaches equilibrium at temperature \(T_{f}\) is given by \[ S_{f} = m c \ln T_{f} \]

Use Entropy Change Formula

The entropy produced is the difference between the final entropy and the initial entropy. Calculate it using the integral form of the entropy change for a temperature gradient.

Derive and Simplify

Substitute the expressions for initial and final entropies into the formula and simplify to get: \[ \sigma = m c \left(1 + \ln T_{f} + \frac{T_{C}}{T_{H} - T_{C}} \ln T_{C} - \frac{T_{H}}{T_{H} - T_{C}} \ln T_{H} \right) \]

Key concepts

Temperature Equilibrium

Temperature equilibrium is achieved when the temperature within a system becomes uniform and no heat flows within it. In the given exercise, the cylindrical rod initially has different temperatures at its ends, but once it is insulated, it reaches a steady final temperature, denoted as \(T_f\). This final equilibrium temperature can be found by taking the average of the initial temperatures at the two ends of the rod:
\[ T_{f} = \frac{T_{H} + T_{C}}{2} \] \[ T_H \] is the higher temperature end, and \[ T_C \] is the lower temperature end. Once this average temperature is achieved throughout the rod, the rod is in thermal equilibrium, and no further temperature changes will happen.
Achieving equilibrium is crucial in studying heat transfer and understanding how systems naturally progress towards a state of uniform temperature over time.

Specific Heat Capacity

Specific heat capacity is a property of materials that indicates how much heat energy is required to change the temperature of a given mass by one degree Celsius (or Kelvin). It is often denoted by the symbol \'c\'. In the context of the given problem, the rod's specific heat capacity \(c\) is used to calculate the entropy change.
The heat energy associated with a temperature change in the rod can be expressed as:
\[ Q = m \cdot c \cdot \Delta T \]
where \[ Q \] is the heat energy, \[ m \] is the mass of the rod, and \[ \Delta T \] is the change in temperature.
Understanding specific heat capacity helps in determining how much energy is stored or released when a material undergoes a temperature change, which is crucial for calculations involving entropy and energy balance in thermodynamic processes.

Entropy Change

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, entropy change is essential to understanding how energy spreads out within a system. For the given exercise, the entropy produced during the process is due to the temperature gradient and the subsequent heat transfer until equilibrium is reached.
Initially, the rod has a linear temperature distribution given by:
\[ T(z) = T_{H} - \left( \frac{T_{H} - T_{C}}{L} \right) z \]
To find the total entropy change, first calculate the initial and final entropy of the rod. The initial entropy isn't uniform due to varying temperature, while the final entropy at equilibrium temperature \[ T_f \] is simpler:
\[ S_{f} = m c \ln T_f \]
The total entropy change produced \[ \sigma \] in the system is given by:
\[ \sigma = m c \left(1 + \ln T_f + \frac{T_C}{T_H - T_C} \ln T_C - \frac{T_H}{T_H - T_C} \ln T_H \right) \]
This formula captures how entropy changes as the rod reaches thermal equilibrium from its initial linear temperature distribution. It helps in understanding the irreversibility and energy dispersal associated with the process.