Question

The temperature of an incompressible substance of mass \(m\) and specific heat \(c\) is reduced from \(T_{0}\) to \(T\left(

Short answer

Plot \ \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \ versus \ x = \frac{T}{T_0} \ for \ 0.8 \leq x \leq 1.0.

Step-by-step solution

- Understand the Problem

The goal is to find the minimum theoretical work input required by a refrigeration cycle and then plot a specific ratio involving that work over a certain range of temperatures.

- Define the Minimum Work Input

Using the second law of thermodynamics for an ideal refrigeration process, the minimum work input, \(W_{\min}\), can be derived via the coefficient of performance (COP). For refrigeration, \(COP = \frac{Q_{\text{out}}}{W_{\text{min}}}\), where \(Q_{\text{out}} = m c (T_0 - T)\).

- Apply the Coefficient of Performance

For an ideal refrigeration cycle, \(COP = \frac{T}{T_0 - T}\). Rearranging this formula to find \(W_{\min}\) yields: \[ W_{\min} = m c (T_0 - T) \frac{T_0}{T} \].

- Simplify the Expression

Next, simplify the work expression into the requested form. The equation becomes \[ \frac{W_{\min}}{m c T_0} = \frac{T_0 - T}{T} \].

- Define the Ratio

Let \(x = \frac{T}{T_0}\), then \( \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \).

- Plot the Function

Plot the function \[ \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \] versus \(x\) for \(x \) ranging from 0.8 to 1.0. This can be done using graphing software or by hand on graph paper.

Key concepts

incompressible substance

An incompressible substance is a material whose volume or density does not change significantly with changes in pressure. This property is typical of most liquids.

In our exercise, we assume the substance is incompressible to simplify the analysis. This assumption is crucial because it means the substance's volume and specific heat (denoted as c) remain constant throughout the process. These constants simplify our calculations for determining the minimum theoretical work input.

specific heat

The specific heat of a substance, represented by the symbol c, is the amount of heat required to change the temperature of a unit mass of the substance by one degree. It is defined by the equation:

\[ \text{specific heat} = \frac{Q}{m \times \triangle T} \ \ \] where \(Q\) is the heat added or removed, \(m\) is the mass, and \( \triangle T\) is the change in temperature.

In our context, knowing the specific heat helps us relate the heat transfer needed to achieve a particular temperature change (from \(T_0\) to \(T\)) in the refrigeration cycle. This relationship is critical for calculating the required work input.

second law of thermodynamics

The Second Law of Thermodynamics is fundamental in refrigeration cycle analysis. It states that heat transfer occurs spontaneously from higher to lower temperatures. For refrigeration, we need to input work to transfer heat in the opposite direction.

This principle is key to understand minimum work input using the ideal coefficient of performance (COP). By leveraging the second law, we derive the relationship:

\[ \text{COP} = \frac{Q_{\text{out}}}{W_{\text{min}}} \ \]

Where:

• COP is the coefficient of performance
• \(Q_{\text{out}}\) is the heat removed
• \(W_{\text{min}}\) is the minimum work input

coefficient of performance

The coefficient of performance (COP) is a measure of a refrigeration cycle's efficiency. It is defined as the ratio of useful heating or cooling provided to the work required:

\[ \text{COP} = \frac{T}{T_0 - T} \ \]

For our problem, the equation helps us determine the work required for ideal scenarios. Replacing the COP value in our equations and rearranging gives us:

\[ W_{\text{min}} = m c (T_0 - T) \frac{T_0}{T} \ \ \]
This equation shows how the work required depends on temperature changes and the specific heat capacity of the substance. Simplifying further allows plotting the desired function:

\[ \frac{W_{\text{min}}}{m c T_0} = \frac{T_0 - T}{T} \ \] Letting \( x = \frac{T}{T_0} \), we get: \[ \frac{W_{\text{min}}}{m c T_0} = \frac{1 - x}{x} \ \]