Problem 21
Question
Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process? Explain.
Step-by-Step Solution
Verified Answer
(a) The ice cube can melt reversibly when it is at its melting point (0°C or 273.15 K) and the pressure is constant (at atmospheric pressure).
(b) No, ∆E is not zero during a reversible melting process because both heat is absorbed and work is done by the ice cube, indicating that energy is transferred within the system.
1Step 1: (a) Identifying reversible melting conditions
To allow the ice cube to melt reversibly, the melting process must occur under thermodynamic equilibrium. This means:
1. The temperature of the ice cube should be equal to its melting point (0°C or 273.15 K)
2. The pressure applied on the ice cube must be equal to the atmospheric pressure (around 1 atm or 101325 Pa)
In conclusion, the ice cube can melt reversibly when it is at its melting point and the pressure is constant (at atmospheric pressure).
2Step 2: (b) Determining if ∆E is zero during reversible melting
In order to determine if the change in internal energy (∆E) is zero during a reversible melting process, we can use the first law of thermodynamics, which states:
∆E = Q - W
Where:
- ∆E is the change in internal energy of the system
- Q is the heat added to the system
- W is the work done by the system
During a constant-pressure reversible melting process, the ice cube absorbs heat (Q) at its melting point, which is used to break the hydrogen bonds within the ice. Simultaneously, the volume of the ice cube changes as the ice transforms into liquid water. This change in volume results in work done by the ice cube (W).
Since both heat is absorbed and work is done by the ice cube, it implies that the energy is transferred within the system. Therefore, we can conclude that the change in internal energy (∆E) is NOT zero for the reversible melting process of the ice cube.
Key Concepts
Thermodynamic EquilibriumFirst Law of ThermodynamicsInternal Energy Change
Thermodynamic Equilibrium
When we talk about reversible melting of an ice cube, thermodynamic equilibrium is key. At thermodynamic equilibrium, there are no unbalanced forces within the system. For the ice cube, this equilibrium is reached when its temperature is equal to its melting point, which is 0°C (or 273.15 K). At this temperature, the ice cube can transition between solid and liquid phases without any net energy flow between it and the surrounding environment. Additionally, the pressure must be constant, typically at atmospheric pressure (1 atm or 101325 Pa).
In practice, this means the ice cube remains stable in its boundary conditions, ensuring a reversible process. Reversible processes are theoretically ideal, where the system changes state in an infinitely slow manner so as to maintain equilibrium.
In practice, this means the ice cube remains stable in its boundary conditions, ensuring a reversible process. Reversible processes are theoretically ideal, where the system changes state in an infinitely slow manner so as to maintain equilibrium.
- Temperature equals melting point: 0°C
- Constant pressure: 1 atm
- System changes occur without energy gradient
First Law of Thermodynamics
The First Law of Thermodynamics is a cornerstone in understanding reversible melting. It's essentially an energy balance law, expressed as: \[ \Delta E = Q - W \]where \( \Delta E \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
During the melting of ice, heat (\( Q \)) is absorbed by the ice at a constant temperature, breaking the hydrogen bonds holding the ice structure. This is essential for the phase change from solid to liquid. As the ice melts, a slight expansion occurs because water in its liquid form occupies more volume than ice. This leads to work being done by the system (\( W \)) as the ice pushes against the atmospheric pressure.
Therefore, according to the First Law, the internal energy is changed by the difference between the heat absorbed and the work done, underscoring that energy cannot be created or destroyed, but only transformed.
During the melting of ice, heat (\( Q \)) is absorbed by the ice at a constant temperature, breaking the hydrogen bonds holding the ice structure. This is essential for the phase change from solid to liquid. As the ice melts, a slight expansion occurs because water in its liquid form occupies more volume than ice. This leads to work being done by the system (\( W \)) as the ice pushes against the atmospheric pressure.
Therefore, according to the First Law, the internal energy is changed by the difference between the heat absorbed and the work done, underscoring that energy cannot be created or destroyed, but only transformed.
Internal Energy Change
Understanding the change in internal energy (\( \Delta E \)) is essential for analyzing the melting process. During reversible melting, the internal energy change tells us about the energy transformations happening inside the system.
As the ice cube melts, it absorbs heat. This heat is utilized to overcome the hydrogen bonds within the ice structure, a process necessary for the phase transition to liquid water. Even though heat is efficiently absorbed to facilitate melting without an increase in temperature, the system does perform work. The expansion causes work to be done against the surrounding pressure.
As the ice cube melts, it absorbs heat. This heat is utilized to overcome the hydrogen bonds within the ice structure, a process necessary for the phase transition to liquid water. Even though heat is efficiently absorbed to facilitate melting without an increase in temperature, the system does perform work. The expansion causes work to be done against the surrounding pressure.
- Energy is absorbed as heat \( Q \)
- Work \( W \) is done due to volume expansion
Other exercises in this chapter
Problem 19
Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from \(300 \mathrm{~K}\) to \(200 \mathrm{~K
View solution Problem 20
A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to
View solution Problem 22
Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pres
View solution Problem 23
(a) How can we calculate \(\Delta S\) for an isothermal process? (b) Does \(\Delta S\) for a process depend on the path taken from the initial state to the fina
View solution