Problem 16 - A Modern Course in Statistical Physics

Question

Assume that two vessels of liquid \(\mathrm{He}^{4}\), connected by a very narrow capillary, are maintained at constant temperature; that is, vessel A is held at temperature \(T_{A}\), and vessel B is held at temperature \(T_{\mathrm{B}}\). If an amount of mass, \(\Delta M\), is transferred reversibly from vessel A to vessel B, how much heat must flow out of (into) each vessel? Assume that \(T_{\mathrm{A}}>T_{\mathrm{B}}\).

Step-by-Step Solution

Answer

Heat \(Q_{A} = L \cdot \Delta M\) must flow out of vessel A, while heat \(Q_{B} = L \cdot \Delta M\) must flow into vessel B.

Step 1 - Understand the Physical Setup

Two vessels (A and B) containing liquid helium-4 are connected by a narrow capillary and held at constant temperatures, with vessel A at temperature \(T_{A}\) and vessel B at temperature \(T_{B}\). Given that \(T_{A} > T_{B}\), an amount of mass \(\Delta M\) is transferred reversibly from vessel A to vessel B. The task is to find out how much heat must flow out of (or into) each vessel.

Step 2 - Consider the Heat Flow due to Mass Transfer

When an amount of mass \(\Delta M\) is transferred between the vessels, we need to consider how this affects the energy and heat flow in each vessel. The specific heat at constant volume for liquid helium-4 can be used to estimate the heat changes: \(C_{v}\), where heat \(Q\) is related to the temperature change and mass by the relation \(Q = C_{v} \cdot \Delta T \cdot \Delta M\). However, since each vessel is maintained at a constant temperature, \(\Delta T = 0\). The heat flow within each vessel must compensate for the transfer of energy due to the movement of mass.

Step 3 - Identify Heat Transfer as Internal Energy Exchange

The internal energy change when helium is transferred can be associated with enthalpy changes. For a reversible process and a small mass transfer, the heat transferred (\(Q_{A}\) for vessel A and \(Q_{B}\) for vessel B) can be modeled in terms of enthalpy (\(H\)). The enthalpy change for the transferred mass can be written as \(\Delta H = H_{B} - H_{A}\), where \(H_{A}\) and \(H_{B}\) are the enthalpies at temperatures \(T_{A}\) and \(T_{B}\), respectively.

Step 4 - Use the Relationship of Reversible Heat Transfer

For reversible heat transfer, the relationship \(Q = T \cdot \frac{dS}{dT} \) and the specific heat at constant pressure or volume can provide the needed heat transfer equation. Since we are dealing with isothermal processes within each vessel, the heat transferred must be found from the latent heat of vaporization (or similar applicable latent heat values in liquid helium-4) per unit mass, denoted as \(L\).

Step 5 - Calculate the Heat Flow

Estimate the heat \(Q_{A}\) needed to remove the mass \(\Delta M\) from vessel A and the heat \(Q_{B}\) needed to add the mass \(\Delta M\) to vessel B. Using the latent heat of vaporization \(L\) of the liquid helium-4:\[Q_{A} = L \cdot \Delta M\] (since heat needs to be supplied to evaporate helium from vessel A), and\[Q_{B} = -L \cdot \Delta M\] (since heat needs to be released to condense helium in vessel B).

Step 6 - Conclude the Heat Flow Dynamics

Summarizing the heat flow: Heat \(Q_{A} = L \cdot \Delta M\) must flow out of vessel A, and an equal amount of heat \(Q_{B} = L \cdot \Delta M\) must be absorbed by vessel B (due to the respective latent heat requirements). The signs indicate that vessel A loses heat (outflow), and vessel B gains heat (inflow).

Key Concepts

heat transfer

In thermodynamics, heat transfer is the process of energy moving from one place to another due to a temperature difference. When looking at liquid helium-4 in this exercise, we see heat transfer when the mass moves between two vessels. The vessels in the problem are kept at different temperatures: vessel A is warmer than vessel B.

Heat always moves from the hotter object to the cooler one. Thus, as mass \(\text{●} M\) moves from vessel A to vessel B, energy in the form of heat flows as well. To understand how much heat flows in or out, we consider specific properties of helium-4, such as its heat of vaporization and temperature conditions in the vessels.

latent heat of vaporization

The latent heat of vaporization is the heat required for turning a liquid into a gas without changing its temperature. For helium-4, the latent heat is critical in understanding the energy required to move the mass \(\text{ΔM}\) between vessels. When helium moves from vessel A to vessel B, it must first vaporize in vessel A (since \(\text{T_A} > \text{T_B}\)).

Here, the heat needed is \[Q_A = L \times \text{ΔM}\text{,}\] where \(L\) is the latent heat of vaporization. Since heat is absorbed in this process, vessel A loses this energy. Conversely, when the helium condenses in vessel B, the same amount of heat is released, given by \[Q_B = -L \times \text{ΔM} \text{.}\text{,}\] Knowing the latent heat value helps in accurately determining these energy exchanges for the mass transferred.

enthalpy change

Enthalpy is a measure of the total energy within a system, which includes internal energy plus the product of pressure and volume. In this exercise, we leverage enthalpy changes to understand heat flows during mass transfer.
For a small mass \(ΔM\), the enthalpy change \(ΔH\) can be described for both vessels. When mass is taken out from vessel A, the enthalpy reduces by \[ΔH = \text{H}_A \text{.}\text{ ΔM} \text{,}\text{,}\] and when added to vessel B, enthalpy increases by \[ΔH = \text{H}_B \text{.}\text{ ΔM} \text{.}\text{.}\text{.}\text{,}\] Since the process is reversible, the heat transfer correlates directly with this enthalpy change. Calculating it helps to understand the exact amount of heat exchanged due to mass transfer.

isothermal processes

Isothermal processes occur at a constant temperature, which simplifies the heat transfer calculations. In the given exercise, both vessels A and B are maintained at constant temperatures, \(T_A\) and \(T_B\), respectively. This means there is no temperature change in either vessel during the mass transfer.

The key takeaway is that in an isothermal situation, the heat flow must balance the energy required for phase changes, like vaporization or condensation, happening at a constant temperature. This allows us to use the latent heat of vaporization to determine heat amounts, ensuring no additional thermal energy needs are due to temperature changes.

specific heat

Specific heat is the amount of heat required to change the temperature of a substance's unit mass by one-degree Celsius or Kelvin. For liquid helium-4, specific heat helps estimate any indirect heat contributions under different setups. However, in the given problem, there is no temperature change in vessels, thus \(ΔT = 0\).

Although specific heat is constant at volume \((C_v)\) or pressure \((C_p)\) for typical situations, in an isothermal process with no temperature change, it plays a minimal role. Instead, the focus is on latent heat properties. Understanding specific heat is essential, generally, to analyze systems where temperature may change, and it allows more detailed heat calculations in various scenarios.