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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

Liquid water is fed to a boiler at \(24^{\circ} \mathrm{C}\) and 10 bar and is converted at constant pressure to saturated steam. (a) Use the steam tables to calculate \(\Delta \hat{H}(\mathrm{kJ} / \mathrm{kg})\) for this process, and then determine the heat input required to produce \(15,800 \mathrm{m}^{3} / \mathrm{h}\) of steam at the exit conditions. Assume the kinetic energy of the entering liquid is negligible and that steam is discharged through a 15 -cm ID pipe. (b) How would the calculated value of the heat input change if you did not neglect the kinetic energy of the inlet water and if the inner diameter of the steam discharge pipe were \(13 \mathrm{cm}\) (increase, decrease, stay the same, or no way to tell without more information)?

A certain gasoline engine has an efficiency of \(30 \% ;\) that is, it converts into useful work \(30 \%\) of the heat generated by burning a fuel. (a) If the congine consumes \(0.80 \mathrm{L}\) /h of a gasoline with a heating value of \(3.25 \times 10^{4} \mathrm{kJ} / \mathrm{L}\), how much power does it provide? Express the answer both in \(\mathrm{kW}\) and horsepower. (b) Suppose the fuel is changed to include \(10 \%\) ethanol by volume. The heating value of ethanol is approximately \(2.34 \times 10^{4} \mathrm{kJ} / \mathrm{L}\) and volumes of gasoline and ethanol may be assumed additive. At what rate ( \((\text { / } h\) ) does the fuel mixture have to be consumed to produce the same power as gasoline?

Air is heated from \(25^{\circ} \mathrm{C}\) to \(140^{\circ} \mathrm{C}\) prior to entering a combustion furnace. The change in specific enthalpy associated with this transition is \(3349 \mathrm{J} / \mathrm{mol}\). The flow rate of air at the heater outlet is \(1.65 \mathrm{m}^{3} / \mathrm{min}\) and the air pressure at this point is \(122 \mathrm{kPa}\) absolute. (a) Calculate the heat requirement in \(\mathrm{kW}\), assuming ideal-gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible. (b) Would the value of \(\Delta \dot{E}_{k}\) [which was neglected in Part (a)] be positive or negative, or would you need more information to be able to tell? If the latter, what additional information would be needed?

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.

Steam at \(260^{\circ} \mathrm{C}\) and 7.00 bar absolute is expanded through a nozzle to \(200^{\circ} \mathrm{C}\) and 4.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. The specific enthalpy of steam is \(2974 \mathrm{kJ} / \mathrm{kg}\) at \(260^{\circ} \mathrm{C}\) and 7 bar and \(2860 \mathrm{kJ} / \mathrm{kg}\) at \(200^{\circ} \mathrm{C}\) and 4 bar. Use the open-system energy balance to calculate the exit steam velocity.

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