Problem 22 - University Physics with Modern 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 195 kg of 30.0\(^\circ\)C water and attempt to warm it further by pouring in 5.00 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 due to spontaneous heat mixing. (b) Final temperature is 31.75°C. (c) Net entropy change is positive.

Step 1: Analyze the Reversible Nature

Determine whether the process is reversible or irreversible by considering the spontaneous mixing of hot and cold water. Since the temperature difference between the hot boiling water and cooler bath water leads to a spontaneous transfer of heat, and the system cannot return to its initial state without external work, the process is irreversible.

Step 2: Use Energy Conservation for Final Temperature

Apply the principle of energy conservation:Let the mass of the bath water be \(m_1 = 195 \text{ kg}\), its initial temperature \(T_1 = 30.0^\circ C\). The mass of the boiling water is \(m_2 = 5.00 \text{ kg}\), with an initial temperature \(T_2 = 100.0^\circ C\). The specific heat capacity of water is \(c = 4.186 \text{ kJ/kg}\cdot\text{°C}\).The final temperature can be found using:\[ m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0 \]Solve for \(T_f\):\[ 195 \cdot (T_f - 30.0) + 5.00 \cdot (T_f - 100.0) = 0 \]

Step 3: Solve for Final Temperature

Continue solving the equation for the final temperature:Expand the terms:\[ 195T_f - 5850 + 5T_f - 500 = 0 \]Combine like terms:\[ 200T_f - 6350 = 0 \]Solve for \(T_f\):\[ T_f = \frac{6350}{200} = 31.75^\circ C \]

Step 4: Entropy Change for Cold Water

Calculate the entropy change of the bath water (cold):\[ \Delta S_{cold} = m_1 c \ln \left(\frac{T_f}{T_1}\right) \]\[ \Delta S_{cold} = 195 \times 4.186 \times \ln \left(\frac{31.75 + 273.15}{30 + 273.15}\right) \]

Step 5: Entropy Change for Hot Water

Calculate the entropy change of the boiling water (hot):\[ \Delta S_{hot} = m_2 c \ln \left(\frac{T_f}{T_2}\right) \]\[ \Delta S_{hot} = 5 \times 4.186 \times \ln \left(\frac{31.75 + 273.15}{100 + 273.15}\right) \]

Step 6: Determine Net Entropy Change

Combine the entropy changes for cold and hot components:\[ \Delta S_{net} = \Delta S_{cold} + \Delta S_{hot} \]Calculate each term to determine \( \Delta S_{net} \) and check if it is positive, confirming the irreversibility.

Key Concepts

Understanding Energy Conservation in the Context of Thermodynamics

In thermodynamics, energy conservation, also known as the first law of thermodynamics, is a fundamental principle. It states that energy cannot be created or destroyed; it can only be transformed from one form to another.

In the context of mixing water to change its temperature, energy conservation plays a crucial role. When you add boiling water to bath water, the energy from the hot water is transferred to the cooler water.

• The total amount of thermal energy in the system remains constant.
• However, the energy is redistributed between the two water masses.
• The final temperature of the water depends on the initial energy states of the bath and boiling water.

By applying the formula involving specific heat capacity and mass, one can determine the final shared temperature after mixing the two water bodies. The formula used is:\[ m_1 \cdot c \cdot (T_f - T_1) + m_2 \cdot c \cdot (T_f - T_2) = 0 \] where:- \( m_1 \) and \( m_2 \) are the masses of the bath and boiling water, respectively.- \( T_1 \) and \( T_2 \) are their initial temperatures.- \( c \) is the specific heat capacity of water.

Solving for \( T_f \) gives us the equilibrium temperature where energy distribution is balanced.

Recognizing Irreversible Processes in Daily Phenomena

Irreversible processes are quite common in our everyday lives. In thermodynamics, these processes are defined by their inability to be reversed without outside intervention. Consider mixing hot boiling water into cooler bath water.

This process is spontaneous due to the natural movement from a state of higher temperature to a state of lower temperature, trying to reach thermal equilibrium.

• The main feature of an irreversible process is that it increases entropy within a closed system.
• Heat spontaneously flows from the hot to the cool water, and cannot "unmix" or spontaneously flow backwards to restore original temperatures.

Attempting to return to the initial state, with colder bath water and hot boiling water, would require additional energy input. Entropy change in an irreversible process increases the system's disorder, indicating energy dispersal.

Exploring Entropy Change: A Measure of Disorder

Entropy is a measure of disorder or randomness within a thermodynamic system. When mixing hot and cold water, the entropy of the system changes. This is because the system transitions from two distinct temperature bodies to one with a uniform temperature. The entropy change \( \Delta S \) is calculated for both the cold bath water and the hot boiling water.

• For cold water, the change in entropy is calculated as: \[ \Delta S_{cold} = m_1 c \ln \left(\frac{T_f}{T_1}\right) \]
• For the hot water, it is calculated as: \[ \Delta S_{hot} = m_2 c \ln \left(\frac{T_f}{T_2}\right) \]

Combined, the total entropy change \( \Delta S_{net} = \Delta S_{cold} + \Delta S_{hot} \) indicates whether the process is irreversible. If \( \Delta S_{net} \) is positive, it confirms the irreversible nature of the process due to increased randomness and energy dispersal.

The Role of Specific Heat Capacity in Temperature Changes

Specific heat capacity is essential in understanding how different substances respond to heat. It defines the amount of heat per unit mass required to raise the temperature by one degree Celsius.

In water, the specific heat capacity is relatively high at approximately 4.186 kJ/kg·°C. This means a lot of energy is required to change water's temperature, which is why water is such an effective heat reservoir.

• Specific heat capacity helps determine how much heat is transferred when different masses and temperatures of water are mixed.
• In our context, it aids in calculating the final equilibrium temperature after mixing hot and cold water.

By using heat capacity in calculations, we see how thermal energy is distributed through the water's mass, allowing us to find the final temperature reached. This understanding of specific heat capacity helps explain why energy transfers with such a small change in temperature for large bodies of water, making it vital for energy conservation calculations.