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Liquid water at \(30.0^{\circ} \mathrm{C}\) and liquid water at \(90.0^{\circ} \mathrm{C}\) are combined in a ratio \((1 \mathrm{kg} \text { cold water/2 } \mathrm{kg}\) hot water). (a) Use a simple calculation to estimate the final water temperature. For this part, pretend you never heard of energy balances. (b) Now assume a basis of calculation and write a closed-system energy balance for the process, assuming that the mixing is adiabatic. Use the balance to calculate the specific internal energy and hence (from the steam tables) the final temperature of the mixture. What is the percentage difference between your answer and that of Part (a)?

Short Answer

Expert verified
The weighted average (Step 1 result) gives you a quick estimate of the final temperature. The energy balance approach (Step 2 result) gives you a more precise temperature. The percentage difference (Step 3 result) then quantifies the gap between the estimate and the precise value.

Step by step solution

01

Simple Calculation (Part a)

Calculate the weighted average of the two temperatures. As there is twice as much hot water as cold water, it weighs in twice as heavily. The formula for a weighted average is \[ \frac{ \sum ( \text{{weight}} \times \text{{value}} ) }{ \sum \text{{weight}} } \] Apply this formula using weight as quantity of water and value as temperature: \[ T_{avg} = \frac{1 \cdot 30 ^\circ C + 2 \cdot 90 ^\circ C}{1 + 2} \]
02

Energy Balance Calculation (Part b)

For an adiabatic process in a closed system, energy conservation implies that the internal energy before mixing is equal to the internal energy after mixing. As mass \times specific heat \times temperature equates to internal energy, set up the equation: \[ (m_{c} \cdot c \cdot T_{c}) + (m_{h} \cdot c \cdot T_{h}) = m_{total} \cdot c \cdot T_{final} \] where \( m_{c} = 1kg \), \( c = 4.186J/(kgK) \), \( T_{c} = 30 ^\circ C \), \( m_{h} = 2kg \), \( T_{h} = 90 ^\circ C \) and \( m_{total} = 3kg \). Solve for \( T_{final} \)
03

Percentage Difference Calculation

Calculate the percentage difference between the final temperature calculated in Part a and Part b using the formula: \[ \% Difference = \frac{ |T_{avg} - T_{final}| }{ T_{avg} } \times 100 \% \]

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Key Concepts

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

Thermodynamics
At its core, thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In practical terms, it considers how energy is converted from one form to another and how it affects matter. For chemical processes, thermodynamics helps us understand how heat transfer, energy transformations and material properties are interconnected.

For example, when mixing two liquids at different temperatures, thermodynamics principles guide us in predicting the final temperature. Key concepts such as heat capacity, which is the quantity of heat needed to change a substance's temperature by a certain amount, play a significant role in such calculations. In studying thermodynamics, students learn to set up energy balances around systems to predict changes due to energy exchanges.
Adiabatic Process
An adiabatic process is a type of thermodynamic process where no heat is exchanged with the surroundings. In other words, the system is perfectly insulated, thus all the energy changes in the system are the result of work done on or by the system.

When we talk about combining two masses of liquid water at different temperatures in an adiabatic mixing process, we're saying there's no heat lost to the environment. Thus, whatever energy the water has before the mixing (in the form of internal energy because of their temperatures) will be conserved throughout the mixing process. Understanding adiabatic processes is fundamental in assessing many real-world engineered systems, such as refrigeration cycles and insulated reaction vessels, where the assumption of no heat loss simplifies the energy balance equations.
Specific Internal Energy
The term specific internal energy refers to the energy contained within a substance due to its temperature, independent of its external conditions like pressure or volume. It's measured per unit mass, typically expressed in units like joules per kilogram (J/kg).

Internal energy includes the kinetic energy of molecules moving and vibrating, plus potential energy due to molecular interactions. For liquids and solids, we often simplify calculations by considering only sensible energy, which relates to the material's temperature, and we use specific heat capacity to relate temperature changes to energy changes. In the provided exercise, specific internal energy is central to setting up the energy balance for the adiabatic mixing process.
Closed-System Energy Balance
A closed-system energy balance is an application of the conservation of energy principle to a system where mass does not cross the system boundary. In our scenario, it means that the energy before mixing (the sum of the specific internal energies of the hot and cold water) must equal the energy after mixing.

In performing a closed-system energy balance, it's assumed that all energy changes within the system are due to the properties and quantities of the substances involved. The step-by-step solution provided for the exercise demonstrates how one would apply the concept of internal energy and use specific heat capacities to account for temperature changes in an adiabatic, closed system. This kind of balance is useful in predicting final temperatures and states of substances in thermochemical processes.

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Most popular questions from this chapter

If a system expands in volume by an amount \(\Delta V\left(\mathrm{m}^{3}\right)\) against a constant restraining pressure \(P\left(\mathrm{N} / \mathrm{m}^{2}\right),\) a quantity \(P \Delta V(\mathrm{J})\) of energy is transferred as expansion work from the system to its surroundings. Suppose that the following four conditions are satisfied for a closed system: (a) the system expands against a constant pressure (so that \(\Delta P=0\) ); (b) \(\Delta E_{\mathrm{k}}=0 ;\) (c) \(\Delta E_{\mathrm{p}}=0 ;\) and (d) the only work done by or on the system is expansion work. Prove that under these conditions, the energy balance simplifies to \(Q=\Delta H\)

Steam produced in a boiler is frequently "wet"-that is, it is a mist composed of saturated water vapor and entrained liquid droplets. The quality of a wet steam is defined as the fraction of the mixture by mass that is vapor. A wet steam at a pressure of 5.0 bar with a quality of 0.85 is isothermally "dried" by evaporating the entrained liquid. The flow rate of the dried steam is \(52.5 \mathrm{m}^{3} / \mathrm{h}\). (a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry steams, and the total mass flow rate of the process stream. (b) Calculate the heat input (kW) required for the evaporation process. (c) Suppose leaks developed in the feed pipe to the dryer and in the dryer exit pipe. Speculate on what you would see at each location.

A Thomas flowmeter is a device in which heat is transferred at a measured rate from an electric coil to a flowing fluid, and the flow rate of the stream is calculated from the measured increase of the fluid temperature. Suppose a device of this sort is inserted in a stream of nitrogen, the current through the heating coil is adjusted until the wattmeter reads \(1.25 \mathrm{kW},\) and the stream temperature goes from \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) before the heater to \(34^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) after the heater. (a) If the specific enthalpy of nitrogen is given by the formula \(\hat{H}(\mathrm{kJ} / \mathrm{kg})=1.04\left[T\left(^{\circ} \mathrm{C}\right)-25\right]\) what is the volumetric flow rate of the gas (L/s) upstream of the heater (i.e., at \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) )? (b) List several assumptions made in the calculation of Part (a) that could lead to errors in the calculated flow rate.

Liquid ethanol is pumped from a large storage tank through a 1 -inch ID pipe at a rate of 3.00 gal/min. (a) At what rate in (i) \(\mathrm{ft} \cdot \mathrm{lb}_{\mathrm{f}} / \mathrm{s}\) and (ii) hp is kinetic energy being transported by the ethanol in the pipe? (b) The electrical power input to the pump transporting the ethanol must be greater than the amount you calculated in Part (a). What would you guess becomes of the additional energy? (There are several possible answers.)

A fuel oil is burned with air in a boiler furnace. The combustion produces \(813 \mathrm{kW}\) of thermal energy, of which \(65 \%\) is transferred as heat to boiler tubes that pass through the furnace. The combustion products pass from the furnace to a stack at \(550^{\circ} \mathrm{C}\). Water enters the boiler tubes as a liquid at \(30^{\circ} \mathrm{C}\) and leaves the tubes as saturated steam at 20 bar absolute. (a) Calculate the rate ( \(\mathrm{kg} / \mathrm{h}\) ) at which steam is produced. (b) Use the steam tables to estimate the volumetric flow rate of the steam produced. (c) Repeat the calculation of Part (b), only assume ideal-gas behavior instead of using the steam tables. Would you have more confidence in the estimate of Part (b) or Part (c)? Explain. (d) What happened to the \(35 \%\) of the thermal energy released by the combustion that did not go to produce the steam?

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