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Write and simplify the closed-system energy balance (Equation \(7.3-4\) ) for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. The solution of Part (a) is given as an illustration. (a) The contents of a closed flask are heated from \(25^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). (b) A tray filled with water at \(20^{\circ} \mathrm{C}\) is put into a freezer. The water tums into ice at \(-5^{\circ} \mathrm{C}\). (Note: When a substance expandsit does work on its surroundings and when it contracts the surroundings do work on it.) (c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container. (d) Repeat Part (c), only suppose that the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically, the temperature in the reactor increased.

Short Answer

Expert verified
The energy balance formulas for the 3 scenarios (b, c, and d) are: (b) \( \Delta U = -Q - W \), (c) \( \Delta U = 0 \), and (d) \( 0= Q - W \)

Step by step solution

01

Problem (b)

Define the system: The system is the tray filled with water. Start by writing down the general closed system energy balance equation:\[ \Delta U= Q - W\]where \( \Delta U \) is the change in internal energy, Q is the heat transferred into the system and W is the work done by the system on the surrounding.When the water freezes, the heat is transferred out of the system (be it to the surroundings or the freezer interiors). Hence the heat transfer \( Q \) is negative. Since water is contracting when it freezes, the surroundings do work on the it, thus the work done \( W \) is also negative.So, the energy balance equation for this system becomes\[ \Delta U = -Q - W\]
02

Problem(c)

Define the system: The system is the chemical reaction taking place in a closed adiabatic rigid container. For an adiabatic process, no heat is transferred, so \( Q = 0 \).For a rigid container, no work is done as there is no change in volume (or expansion or contraction), hence \( W = 0 \).So, the energy balance equation for this system becomes:\[ \Delta U = 0 \]
03

Problem (d)

This problem repeats the same system as in part (c), but the reactor is isothermal rather than adiabatic. In isothermal processes, the temperature remains constant.Recall that the change in internal energy (∆U) is related to the temperature change. If the temperature does not change (as in isothermal processes), there is no change in internal energy: \( \Delta U = 0 \).In this case, however, the reaction is not adiabatic because it allows heat exchange with the surroundings. We do not know whether the heat transfer is into or out of the system or the amount of work done, so we keep \( Q \) and \( W \) in the equation. The energy balance equation for this system becomes: \[ 0= Q - W\]

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. It encompasses several laws that govern energy conversions and the direction that heat transfers. One of the fundamental concepts in thermodynamics is the closed-system energy balance, which describes how the energy within an isolated system changes over time. A closed system is one that can exchange energy (but not matter) with its surroundings. The change in the system's internal energy (abla U) can be expressed as the heat (Q) added to the system minus the work (W) done by the system on its surroundings:[ U = Q - W]The signs of Q and W indicate the direction of heat transfer and work, respectively. If heat is added to the system, Q is positive, and if work is done by the system (such as expansion), W is positive. Conversely, if heat is removed, Q is negative, and if work is done on the system (such as compression), W is negative. Understanding these principles helps us analyze various thermodynamic processes.
Heat Transfer
Heat transfer is a key concept in thermodynamics and involves the movement of thermal energy from one place to another or from one substance to another. It can occur in three primary forms: conduction (through solid objects), convection (through fluids), and radiation (through electromagnetic waves). When examining systems, such as the cooling tray of water in the freezer, understanding that heat transfer is typically from a system to its surroundings helps us determine the appropriate signs for the heat term in closed-system energy balances. A negative Q value implies that the system is losing heat, which is the case when water turns into ice. With processes involving heat transfer, knowing whether a process is adding or removing heat from a system is crucial for predicting changes in properties like temperature and state.
Adiabatic Process
An adiabatic process is a thermodynamic process in which no heat is transferred to or from the working substance. This type of process can occur naturally if a system is perfectly insulated from its surroundings or intentionally in applications where rapid changes prevent heat transfer. In an adiabatic process within a closed system, since Q = 0, the energy balance simplifies to [ U = - W]. However, if the closed system is also rigid, which means no work is done because the volume doesn't change (W = 0), the change in internal energy must also be zero (U = 0), as seen in the solution for the chemical reaction occurring in an adiabatic rigid container. Remembering that adiabatic equates to no heat transfer helps in quickly identifying how the energy balance equation should be modified for such scenarios.
Isothermal Process
An isothermal process is characterized by a constant temperature throughout the process. In these cases, the internal energy does not change because the internal energy of an ideal gas depends only on its temperature. Given that U = 0 for an isothermal process—but not necessarily adiabatic—there can be heat transfer as long as it is balanced by an equal amount of work done. For an isothermal process in a closed system, the energy balance can be represented by [ 0 = Q - W ]. The isothermal condition means that any heat entering or leaving the system must be perfectly compensated by work done on or by the system. This balance is essential in processes like the isothermal reaction in a reactor cited in the exercise, where the heat generated (or absorbed) by the reaction is countered by heat transfer with the surroundings to keep the temperature steady.

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Most popular questions from this chapter

Methane enters a 3 -cm ID pipe at \(30^{\circ} \mathrm{C}\) and 10 bar with an average velocity of \(5.00 \mathrm{m} / \mathrm{s}\) and emerges at a point 200 m lower than the inlet at \(30^{\circ} \mathrm{C}\) and 9 bar. (a) Without doing any calculations, predict the signs ( \(+\) or \(-\) ) of \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}},\) where \(\Delta\) signifies (outlet - inlet). Briefly explain your reasoning. (b) Calculate \(\Delta \dot{E}_{\mathrm{k}}\) and \(\Delta \dot{E}_{\mathrm{p}}(\mathrm{W}),\) assuming that the methane behaves as an ideal gas. (c) If you determine that \(\Delta \dot{E}_{\mathrm{k}} \neq-\Delta \dot{E}_{\mathrm{p}},\) explain how that result is possible.

Superheated steam at 40 bar absolute and \(500^{\circ} \mathrm{C}\) flows at a rate of \(250 \mathrm{kg} / \mathrm{min}\) to an adiabatic turbine, where it expands to 5 bar. The turbine develops \(1500 \mathrm{kW}\). From the turbine the steam flows to a heater, where it is reheated isobarically to its initial temperature. Neglect kinetic energy changes. (a) Write an energy balance on the turbine and use it to determine the outlet stream temperature. (b) Write an energy balance on the heater and use it to determine the required input (kW) to the steam. (c) Verify that an overall energy balance on the two-unit process is satisfied. (d) Suppose the turbine inlet and outlet pipes both have diameters of 0.5 meter. Show that it is reasonable to neglect the change in kinetic energy for this unit.

Saturated steam at \(100^{\circ} \mathrm{C}\) is heated to \(350^{\circ} \mathrm{C}\). Use the steam tables to determine (a) the required heat input (J/s) if a continuous stream flowing at \(100 \mathrm{kg} / \mathrm{s}\) undergoes the process at constant pressure and (b) the required heat input (J) if \(100 \mathrm{kg}\) undergoes the process in a constant-volume container. What is the physical significance of the difference between the numerical values of these two quantities?

Liquid water at 60 bar and \(250^{\circ} \mathrm{C}\) passes through an adiabatic expansion valve, emerging at a pressure \(P_{\mathrm{f}}\) and temperature \(T_{\mathrm{f}} .\) If \(P_{\mathrm{f}}\) is low enough, some of the liquid evaporates. (a) If \(P_{\mathrm{f}}=1.0\) bar, determine the temperature of the final mixture \(\left(T_{\mathrm{f}}\right)\) and the fraction of the liquid feed that evaporates \(\left(y_{\mathrm{v}}\right)\) by writing an energy balance about the valve and neglecting \(\Delta \dot{E}_{\mathrm{k}}\) (b) If you took \(\Delta \dot{E}_{\mathrm{k}}\) into account in Part (a), how would the calculated outlet temperature compare with the value you determined? What about the calculated value of \(y_{\mathrm{v}} ?\) Explain. (c) What is the value of \(P_{\mathrm{f}}\) above which no evaporation would occur? (d) Sketch the shapes of plots of \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\) for 1 bar \(\leq P_{\mathrm{f}} \leq 60\) bar. Briefly explain your reasoning.

A piston-fitted cylinder with a 6 -cm inner diameter contains \(1.40 \mathrm{g}\) of nitrogen. The mass of the piston is 4.50 kg, and a 25.00-kg weight rests on the piston. The gas temperature is 30^ C, and the pressure outside the cylinder is 2.50 atm. (a) Prove that the absolute pressure of the gas in the cylinder is \(3.55 \times 10^{5} \mathrm{Pa}\). Then calculate the volume occupied by the gas, assuming ideal- gas behavior. (b) Suppose the weight is abruptly lifted and the piston rises to a new equilibrium position. Further suppose that the process takes place in two steps: a rapid step in which a negligible amount of heat is exchanged with the surroundings, followed by a slow step in which the gas returns to \(30^{\circ} \mathrm{C}\). Considering the gas as the system, write the energy balances for step \(1,\) step \(2,\) and the overall process. In all cases, neglect \(\Delta E_{\mathrm{k}}\) and \(\Delta E_{\mathrm{p}} .\) If \(\tilde{U}\) varies proportionally with \(T\), does the gas temperature increase or decrease in step 1? Briefly explain your answer. (c) The work done by the gas equals the restraining force (the weight of the piston plus the force due to atmospheric pressure) times the distance traveled by the piston. Calculate this quantity and use it to determine the heat transferred to or from (state which) the surroundings during the process.

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