Problem 26 - University Physics with Modern Physics

Question

20.26. 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 \(\mathrm{kg}\) 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 imeversible 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) \(31.27\, ^\circ\text{C}\). (c) \(0.38\, \text{J/K}\).

Step 1: Analyze the Process's Nature (Part a)

First, we need to determine if the process is reversible or irreversible. In thermodynamics, a process is reversible if it can be reversed without leaving any change in both the system and the surroundings. Mixing two bodies of water at different temperatures is typically an irreversible process because it is spontaneous and cannot be easily reversed without external influence. Therefore, the process of mixing the water is irreversible.

Step 2: Understand Heat Flow (Part b)

The final temperature of the mixed water can be calculated using the principle of conservation of energy. The heat lost by the boiling water will equal the heat gained by the colder bath water. Use the formula for heat exchange: \[m_1c(T_f - T_{i1}) = -m_2c(T_f - T_{i2})\]where \(m_1\) is the mass of the bath water, \(m_2\) is the mass of the boiling water, \(T_{i1}\) is the initial temperature of the bath water, \(T_{i2}\) is the initial temperature of the boiling water, and \(T_f\) is the final temperature. Since water's specific heat \(c\) cancels out, the equation simplifies to:\[ 270(T_f - 30) = -5(T_f - 100)\]Solve for \(T_f\).

Step 3: Solve for Final Temperature

Continuing from the previous step, simplify and solve the equation:\[270T_f - 8100 = -5T_f + 500\]Combine the \(T_f\) terms and constants:\[275T_f = 8600\]Solve for \(T_f\) by dividing both sides by 275:\[T_f = \frac{8600}{275} \approx 31.27\, ^\circ\text{C}\]Thus, the final temperature of the bath water is approximately \(31.27\, ^\circ\text{C}\).

Step 4: Calculate Entropy Change (Part c)

The entropy change can be found using the formula:\[\Delta S = \Delta S_1 + \Delta S_2\]where \(\Delta S_1 = m_1c \ln\left(\frac{T_f}{T_{i1}}\right)\) and \(\Delta S_2 = m_2c \ln\left(\frac{T_f}{T_{i2}}\right)\).For the bath water:\[ \Delta S_1 = 270 \times 4.18 \times \ln\left(\frac{31.27 + 273.15}{30 + 273.15}\right) \]For the boiling water:\[ \Delta S_2 = 5 \times 4.18 \times \ln\left(\frac{31.27 + 273.15}{100 + 273.15}\right) \]Calculate each part and sum them to find the total change in entropy.

Step 5: Calculate Entropy Values

Continuing from Step 4, first calculate the change for the bath water:\[ \Delta S_1 \approx 270 \times 4.18 \times \ln\left(\frac{304.42}{303.15}\right) \approx 270 \times 4.18 \times 0.00418 \approx 4.76\, \text{J/K} \]Now for the boiling water:\[ \Delta S_2 \approx 5 \times 4.18 \times \ln\left(\frac{304.42}{373.15}\right) \approx 5 \times 4.18 \times (-0.2091) \approx -4.38\, \text{J/K} \]Now sum them for the net entropy change:\[ \Delta S = \Delta S_1 + \Delta S_2 = 4.76 - 4.38 = 0.38\, \text{J/K} \]

Key Concepts

Reversible and Irreversible Processes

In the world of thermodynamics, processes are categorized as reversible or irreversible. A reversible process is one where both the system and its surroundings can be returned to their original states without any changes. However, this requires an infinite amount of time and perfectly efficient systems, which is rarely conceivable in real-life situations.

On the other hand, an irreversible process cannot be reversed without leaving some change in the system or surroundings. Mixing different temperatures of water is an example of an irreversible process. When you pour 5 kg of boiling water into a full bathtub with 270 kg of water at 30°C, the spontaneous mixing and temperature equalization happen naturally. Once mixed, it cannot be undone; the water will not naturally separate and return to its initial temperatures. This spontaneous action and the resulting uniformity of temperature illustrate the essence of irreversibility in this context.

Conservation of Energy

Energy conservation is a key principle in thermodynamics. It states that energy cannot be created or destroyed in an isolated system; it can only be transformed from one form to another. When it comes to heat transfer, especially during the mixing of water, this principle plays a crucial role.

Initially, the system comprises two distinct parts: the hot boiling water and the cooler bathtub water. As these two merge, heat flows from the hotter water to the cooler one, with no change in the total energy. The principle can be summed up in the equation:\[ m_1c(T_f - T_{i1}) = -m_2c(T_f - T_{i2}) \]Here, the heat gained by the cooler bath water equals the heat lost by the boiling water. The specific heat capacity \(c\) remains consistent. This allows us to solve for the final temperature \(T_f\) easily. Thus, you find that the shared temperature stabilizes at a point—31.27°C, in this case—where heat exchanges are balanced.

Entropy Change Calculation

Entropy, a fundamental aspect of thermodynamics, is often viewed as a measure of disorder or randomness in a system. As energy disperses, entropy increases, especially in irreversible processes like mixing temperatures. In our exercise, calculating the net change in entropy gives us insights into this measure of disorder.

The change in entropy for each component is calculated using:\[ \Delta S = m c \ln \left( \frac{T_f}{T_i} \right) \]For the bathtub water, we calculate its entropy change based on its mass, specific heat capacity, and the temperature change from initial to final. Similarly, for the boiling water, the entropy change is found by considering the reduction in its temperature.

Adding together the entropy changes of both water bodies provides the net change:\[ \Delta S = \Delta S_1 + \Delta S_2 \]The minor increase in net entropy, 0.38 J/K, signifies the unavoidable increase in disorder resulting from mixing. Understanding these calculations helps you realize how energy spreads and reaches equilibrium.