/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Molten sodium chloride is to be ... [FREE SOLUTION] | 91Ó°ÊÓ

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Molten sodium chloride is to be used as a constant-temperature bath for a high-temperature chemical reactor. Two hundred kilograms of solid \(\mathrm{NaCl}\) at \(300 \mathrm{K}\) is charged into an insulated vessel, and a 3000 kW electrical heater is turned on, raising the salt to its melting point of 1073 K and melting it at a constant pressure of 1 atm. (a) The heat capacity \(\left(C_{p}\right)\) of solid \(\mathrm{NaCl}\) is \(50.41 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) at \(T=300 \mathrm{K},\) and \(53.94 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) at \(T=500 \mathrm{K},\) and the heat of fusion of \(\mathrm{NaCl}\) at \(1073 \mathrm{K}\) is \(30.21 \mathrm{kJ} / \mathrm{mol} .\) Use these data to determine a linear expression for \(C_{p}(T)\) and to calculate \(\Delta \hat{H}\) ( \(\mathrm{kJ} / \mathrm{mol}\) ) for the transition of \(\mathrm{NaCl}\) from a solid at 300 K to a liquid at \(1073 \mathrm{K}\). (b) Write and solve the energy balance equation for this closed system isobaric process to determine the required heat input in kilojoules. (c) If \(85 \%\) of the full power of \(3000 \mathrm{kW}\) goes into heating and melting the salt, how long does the process take?

Short Answer

Expert verified
The linear expression for heat capacity, \(C_{p}(T)\), is \(0.01765T + 44.605 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\). The total heat change, \(\Delta \hat{H}\), for the transition from solid NaCl at 300K to liquid at 1073K is 83.15 kJ/mol. The required heat input for the process is 285000 kJ, and it takes around 112 hours to complete the process.

Step by step solution

01

Developing the linear expression \(C_{p}(T)\)

Given \(C_{p}\) values at 300K and 500K, the equation of a line can be used to find the linear expression for \(C_{p}(T)\). The formula is \(y = mx + b\), where \(m\) (slope) is \(\Delta y/\Delta x\) and \(b\) (intercept) is \(y - mx\). Here, \(y\) represents \(C_{p}\), \(x\) represents \(T\), and \(m\) is \((53.94 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) - 50.41 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}))/(500 \mathrm{K} - 300 \mathrm{K}) = 0.01765 \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}^{2})\). The y-intercept \(b\) can be derived with \(y - mx\) using the \(C_{p}\) and \(T\) values at 300K: \(C_{p} - mT = b => 50.41 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) - (0.01765 \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}^{2})) \cdot 300 \mathrm{K} = 44.605 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\). Therefore, the equation of \(C_{p}(T)\) is \(0.01765T + 44.605 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\).
02

Calculating the enthalpy change \(\Delta \hat{H}\)

Using the developed equation from step 1, \(\Delta \hat{H}\) can be calculated in two parts: the heat gained to increase the temperature (\(\Delta H_{1}\)) and the heat given at the phase change (\(\Delta H_{2}\)). \(\Delta H_{1} = \int_{300}^{1073} C_{p}(T) dT = \int_{300}^{1073} (0.01765T + 44.605) dT\) which can be evaluated as \(\Delta H_{1} = 52.94 \mathrm{kJ/mol}\). \(\Delta H_{2}\) is given as the heat of fusion at 1073K, which is 30.21 kJ/mol. \(\Delta \hat{H}\) is the sum of \(\Delta H_{1}\) and \(\Delta H_{2}\), which is 83.15 kJ/mol.
03

Determining the heat input

Based on the energy balance equation \(Q = \Delta U + W\), and accounting for the process being at constant pressure (isobaric) and in a closed system (no work done), the heat input required (\(Q\)) is equal to \(\Delta H\) (enthalpy change). Given 200kg of NaCl, we can find the molar quantity by dividing by the molar mass of NaCl (58.44 g/mol). Then, multiply it with \(\Delta \hat{H}\) (molar enthalpy change) to get \(Q\), which is \(285000 \mathrm{kJ}\).
04

Computing process time

The time required for the heating process can be calculated by using the formula \(time = Energy/Power\). Knowing that 85% of the full power is used for the process, the actual power used is 0.85 x 3000kW = 2550kW. Therefore, the time required will be \(285000 \mathrm{kJ} /2550 \mathrm{kW} = 112 \mathrm{hours}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Capacity
In thermodynamics, heat capacity is a crucial concept that describes the amount of heat energy required to raise the temperature of a substance by a specific amount. It is often represented as \(C_p\) when measurements are taken at constant pressure. The heat capacity is given in units of \(\text{J/(mol}\cdot\text{K)}\), demonstrating the energy needed to heat one mole of a substance by one Kelvin.

Key factors to remember about heat capacity include:
  • It varies with temperature; hence, it's important to obtain values at different temperatures, as demonstrated with NaCl at 300 K and 500 K.
  • In this exercise, the linear relationship of \(C_p(T)\) was established using the provided data, allowing us to predict \(C_p\) across temperatures.
The linear expression related to heat capacity for NaCl was derived using the formula \(y = mx + b\), where the slope \(m\) captured the change of heat capacity with temperature. Understanding this provides insight into the energy needed for heating processes in practical applications.
Enthalpy Change
Enthalpy change, denoted as \(\Delta \hat{H}\), represents the total heat content change of a system under constant pressure. It's a vital concept to comprehend the energy transactions during chemical processes, particularly those involving phase transitions.

In our scenario of molten sodium chloride:
  • Enthalpy change was evaluated in two stages: heating the solid to its melting point and then allowing it to transition into a liquid phase.
  • The calculation combined the integrated heat capacity function over the temperature range and added the heat of fusion at 1073 K.
  • Mathematically, this involved evaluating \(\int_{300}^{1073} C_p(T)\, dT\) and adding the latent heat of fusion, quantified here as \(\Delta H_2 = 30.21 \, \text{kJ/mol}\).
Gaining an understanding of enthalpy change enhances one's ability to predict how much heat energy a process will need, which is crucial for designing and operating thermal systems.
Energy Balance Equation
In order to thoroughly analyze a thermodynamic system, one must employ the energy balance equation. For an isobaric process like the melting of sodium chloride, the energy equation simplifies as the work done is zero (since the system is closed and pressure is constant).

The equation used was \(Q = \Delta U + W\), where \(Q\) is the heat added to the system, \(\Delta U\) is the change in internal energy, and \(W\) is work done. For this type of process, since work done \(W = 0\), the heat input \(Q\) is directly equal to the enthalpy change \(\Delta H\).

Steps involved include:
  • Calculating the total mole quantity using NaCl's molar mass, essential for determining the entire heat input with \(\Delta \hat{H}\).
  • Identifying that the heat input \(Q\) needed for the process is \(285,000 \, \text{kJ}\).
  • Converting the energy needs into time, considering actual power usage (85% efficiency of the heater), to obtain process duration.
Through mastering the energy balance, engineers and chemists can precisely determine the necessary power supply and duration for thermal operation, an essential step in planning large-scale thermal systems.

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Most popular questions from this chapter

Propane gas enters a continuous adiabatic heat exchanger \(^{17}\) at \(40^{\circ} \mathrm{C}\) and \(250 \mathrm{kPa}\) and exits at \(240^{\circ} \mathrm{C}\). Superheated steam at \(300^{\circ} \mathrm{C}\) and 5.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure. (a) Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed \(\left(\mathrm{m}^{3}\right),\) the mass of steam fed \((\mathrm{kg}),\) and the volume of steam fed \(\left(\mathrm{m}^{3}\right)\) (b) Calculate values of the labeled specific enthalpies in the following inlet-outlet enthalpy table for this process. $$\begin{array}{|l|cc|cc|} \hline \text { Species } & n_{\text {in }} & \hat{H}_{\text {in }} & n_{\text {out }} & \hat{H}_{\text {out }} \\ \hline \mathrm{C}_{3} \mathrm{H}_{8} & 100 \mathrm{mol} & \hat{H}_{\mathrm{a}}(\mathrm{kJ} / \mathrm{mol}) & 100 \mathrm{mol} & \hat{H}_{\mathrm{c}}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{H}_{2} \mathrm{O} & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{b}}(\mathrm{kJ} / \mathrm{kg}) & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{d}}(\mathrm{kJ} / \mathrm{kg}) \\ \hline \end{array}$$ (c) Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams ( \(\mathrm{m}^{3}\) steam fed \(/ \mathrm{m}^{3}\) propane fed). Assume ideal-gas behavior for the propane but not the steam and recall that the exchanger is adiabatic. (d) Calculate the heat transferred from the water to the propane ( \(k J / m^{3}\) propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.) (e) Over a period of time, scale builds up on the heat-transfer surface, resulting in a lower rate of heat transfer between the propane and the steam. What changes in the outlet streams would you expect to see as a result of the decreased heat transfer?

An adiabatic membrane separation unit is used to dry (remove water vapor from) a gas mixture containing 10.0 mole \(\% \mathrm{H}_{2} \mathrm{O}(\mathrm{v}), 10.0\) mole \(\% \mathrm{CO},\) and the balance \(\mathrm{CO}_{2} .\) The gas enters the unit at \(30^{\circ} \mathrm{C}\) and flows past a semipermeable membrane. Water vapor permeates through the membrane into an air stream. The dried gas leaves the separator at \(30^{\circ} \mathrm{C}\) containing \(2.0 \mathrm{mole} \% \mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) and the balance \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). Air enters the separator at \(50^{\circ} \mathrm{C}\) with an absolute humidity of \(0.002 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} / \mathrm{kg}\) dry air and leaves at \(48^{\circ} \mathrm{C}\). Negligible quantities of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{N}_{2}\) permeate through the membrane. All gas streams are at approximately 1 atm. (a) Draw and label a flowchart of the process and carry out a degree of freedom analysis to verify that you can determine all unknown quantities on the chart. (b) Calculate (i) the ratio of entering air to entering gas (kg humid air/mol gas) and (ii) the relative humidity of the exiting air. (c) List several desirable properties of the membrane. (Think about more than just what it allows and does not allow to permeate.)

Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton's rule and Chen's rule and compare the results with a tabulated value of this quantity. Calculate the percentage error that results from using each estimation. Then estimate \(\Delta \hat{H}_{\mathrm{v}}\) at \(100^{\circ} \mathrm{C}\) using Watson's correlation.

The flow of groundwater often plagues construction of tunnels and other underground systems. Dne way of preventing it is with an ice seal-freezing the water in the soil so that the ice formed is a barrier to water movement. Such a structure was planned for the Fukushima TEPCO nuclear power plant, which was severely damaged by a 2011 tsunami that created a tremendous environmental challenge. A major concern was potential contamination with radioactive isotopes of groundwater flowing under the plant and into the ocean. A proposal under consideration was to channel the flow around the plant by forming an ice dam with a 1,400 -meter perimeter, a depth of \(30 \mathrm{m},\) and a thickness of approximately \(2 \mathrm{m}\). This was to be done by pumping a brine solution at a temperature of \(-40^{\circ} \mathrm{C}\) though vertical pipes spaced at \(1-\mathrm{m}\) intervals. The brine would exit at a temperature no greater than \(-25^{\circ} \mathrm{C}\). To keep ambient temperature fluctuations from causing occasional melting, the dam was to have a mean temperature of about \(-20^{\circ} \mathrm{C}\). (a) Estimate the average cooling rate (kW) and associated flow rate of brine (L/min) required to complete formation of the dam within 60 days of starting the refrigeration system. State and give your rationale for each of the assumptions and/or approximations necessary to obtain your result. (b) From a suitable reference, for which you must provide a citation, find an estimate of the ratio of the heat removed to the work done by a refrigeration system. Use the value to estimate the power usage during the time the dam is being created. (c) It is expected that substantially less power will be used once the dam has been formed. Explain. (d) Identify the primary radioactive species that were of greatest concern regarding contamination of the groundwater. (e) Explosions of hydrogen occurred in the power plants after the cooling water system was shut down upon being flooded by the tsunami. What was the source of the hydrogen? Describe the scenario that led to hydrogen formation.

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