Problem 25 - University Physics with Modern Physics

Question

20.25. A sophomore with nothing better to do adds heat to 0.350 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at a temperature of \(25.0^{\circ} \mathrm{C}\) . What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

Step-by-Step Solution

Answer

(a) 428.1 J/K; (b) -392.2 J/K; (c) 35.9 J/K.

Step 1: Understanding the concepts

To solve this problem, we need to understand the concept of entropy change in a system. Entropy is the measure of disorder, and when heat is added at a constant temperature, the change in entropy \( \Delta S \) is given by \( \Delta S = \frac{Q}{T} \), where \( Q \) is the heat added and \( T \) is the absolute temperature in Kelvin.

Step 2: Calculate the heat needed to melt the ice

The heat needed to melt ice is calculated using the formula \( Q = m \cdot L_f \), where \( m = 0.350 \) kg is the mass of the ice and \( L_f = 334,000 \) J/kg is the latent heat of fusion for ice. Substituting the given values, we have \( Q = 0.350 \times 334,000 = 116,900 \) J.

Step 3: Calculate the change in entropy of the water

Once the ice is turned into water at \(0.0^{\circ} \mathrm{C}\), we can calculate its change in entropy using \( \Delta S_{water} = \frac{Q}{T} \). The temperature in Kelvin is \( T = 273.15 \). Therefore, \( \Delta S_{water} = \frac{116,900}{273.15} \approx 428.1 \) J/K.

Step 4: Calculate the change in entropy of the heat source

The massive body at \(25.0^{\circ} \mathrm{C}\), or \(298.15 \) K in Kelvin, loses the same amount of heat \(116,900\) J. Thus, its change in entropy is \( \Delta S_{source} = -\frac{116,900}{298.15} \approx -392.2 \) J/K.

Step 5: Calculate the total change in entropy of the system

The total change in entropy of the system is the sum of the change in entropy of the water and the change in entropy of the heat source: \( \Delta S_{total} = \Delta S_{water} + \Delta S_{source} = 428.1 + (-392.2) = 35.9 \) J/K.

Key Concepts

Latent Heat of Fusion

When ice melts into water, it requires a specific amount of energy known as the latent heat of fusion. This energy is necessary to overcome the forces holding the ice's solid structure together.
For water, the latent heat of fusion is relatively high, at 334,000 Joules per kilogram. This means it takes a significant amount of heat energy to convert ice at 0°C into liquid water at the same temperature, without any temperature change during the process.
Understanding latent heat of fusion is crucial when dealing with phase changes. It is a key concept in calculating the energy needed for these transformations, important for various practical applications.

• The latent heat is unique to each material, reflecting how different substances require varying amounts of energy for phase changes.
• The energy used in overcoming intermolecular forces during melting does not increase the kinetic energy of particles; instead, it breaks the structure apart.
• In the provided exercise, 116,900 J of energy was needed to melt 0.35 kg of ice, all due to the latent heat of fusion for ice.

Heat Transfer

Heat transfer is the process where thermal energy moves from a hotter object to a cooler one. It can occur through various methods such as conduction, convection, and radiation. In this exercise, we focus on how thermal energy was transferred from a massive body at 25°C to ice at 0°C until it melted into water.
When dealing with heat transfer during phase changes, it is crucial to remember that the temperature doesn't necessarily change. Instead, the energy goes into the phase transformation itself.

• Conduction is the primary method of heat transfer when two objects are in direct contact, as seen in this problem.
• It is important to consider the materials' thermal properties like specific heat and latent heat, which determine the energy transfer capabilities.
• In our exercise, 116,900 J was transferred from the warm body to the ice, which helped achieve its phase change without raising the water's temperature.

Entropy Change

Entropy is a concept that quantifies the amount of disorder or randomness in a system. Through processes like melting, the entropy often changes, indicating a higher level of molecular disorder.
In the scenario given, we see a transition of ice to water and heat loss from a heat source, which entails specific changes in entropy. The entropy change is governed by the equation:\[\Delta S = \frac{Q}{T}\]where \( \Delta S \) is the change in entropy, \( Q \) is the heat added or removed, and \( T \) is the absolute temperature in Kelvin.Clarity around entropy changes during heat transfer is essential to understanding thermodynamics. This problem provides an insight into how these calculations are necessary for assessing efficiencies and irreversibility in processes.

• The increase in entropy when ice melts reflects the greater disorder in liquid water compared to solid ice.
• Meanwhile, the heat source experiences an entropy decrease by giving up energy, as observed with a \(-392.2\, \text{J/K}\) change.
• Finally, the overall entropy increases, aligning with the second law of thermodynamics, as seen in the total system change of \(35.9\, \text{J/K}\).