Problem 29 - College Physics

Question

You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 270 kg of \(30.0^{\circ} \mathrm{C}\) water and attempt to warm it further by pouring in 5.00 \(\mathrm{kg}\) of boiling water from the stove. (a) Is this a reversible or an irreversible process? Use physical reasoning to explain. (b) Calculate the final temperature of the bath water. (c) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.

Step-by-Step Solution

Answer

(a) Irreversible; (b) Final temperature: \(31.28^{\circ}C\); (c) Net entropy change: \(5.877 \text{ J/K}\).

Step 1: Understanding the Process

In thermodynamics, a reversible process is an ideal process that never occurs in nature. An irreversible process is actual and results in entropy production because of factors like friction, mixing, or temperature gradients. Here, when hot water is mixed with warm water, the process is irreversible due to the mixing and thermal equilibrium being achieved spontaneously.

Step 2: Calculating Final Temperature

To find the final temperature of the water after mixing, use the principle of conservation of energy (assuming no heat is lost to the environment). Apply the formula for mixing two water masses: \[ m_1 c (T_f - T_1) + m_2 c (T_f - T_2) = 0 \] where \( m_1 = 270 \text{ kg} \), \( T_1 = 30^{\circ}C \), \( m_2 = 5 \text{ kg} \), \( T_2 = 100^{\circ}C \), and \( c = 4.18 \text{ kJ/kg°C} \). Solving gives:\[ 270 \times 4.18 \times (T_f - 30) + 5 \times 4.18 \times (T_f - 100) = 0 \]Solving for \( T_f \):\[ 1131T_f - 33930 + 20.9T_f - 2090 = 0 \]\[ 1151.9T_f = 36020 \]\[ T_f = \frac{36020}{1151.9} \approx 31.28^{\circ}C \]

Step 3: Entropy Change Calculation

Calculate the change in entropy for both masses of water. The change in entropy \( \Delta S \) for water is given by:\[ \Delta S = m c \ln \left( \frac{T_f}{T_i} \right) \]- For the 270 kg of water: \[ \Delta S_1 = 270 \times 4.18 \times \ln \left( \frac{31.28 + 273.15}{30 + 273.15} \right) \]- For the 5 kg of boiling water: \[ \Delta S_2 = 5 \times 4.18 \times \ln \left( \frac{31.28 + 273.15}{100 + 273.15} \right) \]Calculate and add these:\[ \Delta S_1 = 270 \times 4.18 \times \ln \left( \frac{304.43}{303.15} \right) \approx 10.527 \text{ J/K} \]\[ \Delta S_2 = 5 \times 4.18 \times \ln \left( \frac{304.43}{373.15} \right) \approx -4.65 \text{ J/K} \]\[ \Delta S_{total} = \Delta S_1 + \Delta S_2 \approx 10.527 - 4.65 \approx 5.877 \text{ J/K} \]

Step 4: Conclusion

The mixing process is irreversible, the final temperature is approximately \(31.28^{\circ}C\), and the net change in entropy of the system is about \(5.877 \text{ J/K}\).

Key Concepts

Reversible Process

In the realm of thermodynamics, it’s important to understand the concept of a reversible process. This type of process is idealized and theoretical. In a reversible process, the system changes in such a way that the system and its surroundings can be returned to their exact original states by simply reversing the process. Key characteristics include:

• No entropy production or minimum possible increase of entropy.
• Occurs infinitely slowly to allow the system to maintain equilibrium at all times.

It’s virtually impossible to achieve in reality but provides a baseline for understanding efficiency and energy exchanges. The idea helps engineers and scientists model systems to ensure maximum efficiency.

Irreversible Process

Unlike reversible processes, irreversible processes are far more common and occur in real-world applications. These processes are marked by the spontaneous movement towards equilibrium, leading to changes that cannot be simply reversed. An irreversible process is characterized by:

• An increase in entropy, resulting from factors such as mixing or temperature differences.
• Occurs due to natural tendencies such as friction, inelastic deformations, or uncontrolled heat exchange.

Mixing hot and cold water as described in the exercise is a classic example of an irreversible process. The entropy of the system increases as thermal energy disperses, making it impossible to revert to the initial state without external work.

Entropy Change

Entropy is a fundamental concept in thermodynamics, representing the degree of disorder or randomness within a system. When processes occur, entropy typically increases, especially in irreversible processes. This increase is aligned with the second law of thermodynamics, which states that total entropy always increases in a closed system.In the bathwater exercise, calculating the change in entropy involves assessing the energy dispersal when different temperature waters mix. The formula used is:\[ \Delta S = m c \ln \left( \frac{T_f}{T_i} \right) \]This calculation indicates how much disorder is introduced during the mixing. For the individual water masses, different entropy changes were observed, with a net positive outcome, reflecting increased disorder and irreversible nature of the process.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This principle is vital for solving thermodynamics problems, like finding the final temperature during mixing of two water masses.Using the formula:\[ m_1 c (T_f - T_1) + m_2 c (T_f - T_2) = 0 \]we derive the final temperature by equating the heat lost by one body to the heat gained by another, assuming no energy is lost to the surroundings. This application helps ensure that, despite the process being irreversible, the energy within the system remains constant. By conserving energy, we can confidently calculate the resulting equilibrium temperature of the mixed water.