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A linear equation is a mathematical expression that establishes a line when plotted on a graph. In simple terms, it's an equation that describes a straight line relationship between two variables. Typically, it takes the form of \(y = mx + b\), where \(y\) is the dependent variable, \(x\) is the independent variable, \(m\) is the slope of the line, and \(b\) is the y-intercept, defining the point where the line crosses the y-axis.
In the context of specific enthalpy, the exercise provides two points: \((30, 25.8)\) and \((50, 129.8)\). These represent the temperature \(T\) in degrees Celsius and the specific enthalpy \(\hat{H}\) in \(\text{kJ/kg}\), respectively. The goal is to find a linear equation connecting these variables. We calculate the slope \(m\) as the change in enthalpy divided by the change in temperature:

• \(m = \frac{129.8 - 25.8}{50 - 30} = 5.2 \text{kJ/kg°C}\).

Then, by substituting one of the given points into the equation \(\hat{H} = mT + b\), we determine \(b\), the specific enthalpy at the reference temperature:

• \(b = 25.8 - 5.2 \times 30 = -130.2 \text{kJ/kg}\)

Therefore, the linear equation is \(\hat{H} = 5.2T - 130.2\).
This equation now allows for the calculation of specific enthalpy at any temperature within the given range by inserting the value of \(T\) in degrees Celsius.

Heat transfer rate is a measure of how much heat energy is being transferred from one system to another per unit of time. In our case, it can be used to calculate the amount of heat that must be removed from liquid \(n\)-hexane to cool it from one temperature to another.
The heat transfer rate \(Q\) is determined by calculating the difference between initial and final specific enthalpies, multiplied by the mass flow rate \(\dot{m}\). The mass flow rate is given as 20 kg/min, and we convert it to kg/s by dividing by 60: \(\frac{20 \text{ kg/min}}{60 \text{ s/min}} = \frac{1}{3} \text{ kg/s}\). Thus, the formula used is

• \(Q = \dot{m} \times (\hat{H}_{\text{HOT}} - \hat{H}_{\text{COLD}})\)
• \(\hat{H}_{\text{HOT}}\) is the enthalpy at 60°C \((5.2 \times 60 - 130.2 = 182 \text{kJ/kg})\)
• \(\hat{H}_{\text{COLD}}\) is the enthalpy at 25°C \((5.2 \times 25 - 130.2 = -0.2 \text{kJ/kg})\)

By substituting these into the equation,

• \(Q = \frac{1}{3} \times (182 - (-0.2)) = 36.7 \text{kJ/s}\)

This calculation tells us that to cool the \(n\)-hexane from 60°C to 25°C, 36.7 kJ of heat must be transferred out of the system every second.

Internal energy in thermodynamics describes the total energy contained within a system due to both its thermal and chemical states. It includes both the kinetic and potential energies of the molecules within the system. For many practical purposes and under constant pressure conditions, we can approximate specific internal energy \(\hat{U}\) to be equivalent to specific enthalpy \(\hat{H}\).
This approximation holds because the term \(P\hat{V}\), representing the pressure and specific volume product, is often negligible compared to the enthalpy value under typical conditions such as 1 atm.

• Hence, in this exercise, \( \hat{U} = \hat{H} \).

Thus, using the linear equation developed for specific enthalpy, the internal energy can also be expressed as \(\hat{U} = 5.2T - 130.2\).
To understand how internal energy changes with temperature, we compute the difference between the internal energy at two temperatures (60°C and 25°C):

• \(\Delta \hat{U} = (\hat{U}_{\text{HOT}} - \hat{U}_{\text{COLD}})\).
• \(\Delta \hat{U} = (5.2 \times 60 - 130.2) - (5.2 \times 25 - 130.2) = 182 \text{kJ/kg} \).

This value represents the amount of internal energy change when \(n\)-hexane is cooled under the specified conditions, informing us that 182 kJ/kg of internal energy is lost during the cooling process.