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Chapter 10: Problem 27

An electrical coil is used to heat \(20.0 \mathrm{kg}\) of water in a closed well-insulated vessel. The water is initially at \(25^{\circ} \mathrm{C}\) and 1 atm. The coil delivers a steady \(3.50 \mathrm{kW}\) of power to the vessel and its contents. (a) Write a differential energy balance on the water, assuming that \(97 \%\) of the energy delivered by the coil goes into heating the water. What happens to the other \(3 \% ?\) (b) Integrate the equation of Part (a) to derive an expression for the water temperature as a function of time. (c) How long will it take for the water to reach the normal boiling point? Will it boil at this temperature? Explain your answer.

Short Answer

Expert verified
The function of change in temperature over time is \(T(t) = \frac{\dot{W} \cdot f \cdot t}{m \cdot c} + 25\), and the time it takes for the water to reach boiling point can be calculated using \(t = \frac{m \cdot c \cdot (100 - 25)}{\dot{W} \cdot f}\). Although the water reaches the boiling point, it will not transition into steam in the closed, pressurized situation.

Step by step solution

01

Write energy differential equation

Define \(\Delta Q\) as the energy received by the water which equals to \(\dot{W} \cdot f \cdot dt\), where \( \dot{W}\) is the power produced by the coil, \(f\) is the percentage of the energy being used to heat the water and \(dt\) is the very small period of time. Also, energy transferred to a mass \(\Delta Q\) can be expressed as \(m \cdot c \cdot \Delta T\), where \( m \) is mass of the mass, \(c\) is the specific heat capacity of water, therefore, \( \Delta T = \frac{\dot{W} \cdot f \cdot dt }{m \cdot c} \). The 3% of energy delivered by the coil that does not go to heating the water most likely will be dissipated as loss to surrounding through radiation or air currents or sound.
02

Integrate to find Temperature as a function of time

Integrating the equation from Step 1 over the time period from 0 to \(t\) will give the change in the temperature as a function of time. \(\Delta T(t) = \int_{25}^{T}dT = \int_{0}^{t}\frac{\dot{W} \cdot f \cdot dt}{m \cdot c}\). Solving this equation gives: \(T(t) - 25 = \frac{\dot{W} \cdot f \cdot t}{m \cdot c}\), or \(T(t) = \frac{\dot{W} \cdot f \cdot t}{m \cdot c} + 25\)
03

Calculate the time it takes water to reach boiling point

Substitute \(100 ^\circ C\) to the temperature into the formula from Step 2 which will yield the time it takes for the water to boil. \(t = \frac{m \cdot c \cdot (100 - 25)}{\dot{W} \cdot f}\). Then, calculate the time given the values of power, mass, specific heat and efficiency.
04

Discuss if the water will boil at the reached temperature

Although the water reaches boiling point, it will not boil until the pressure above the water changes. In a closed vessel, water cannot transition to steam because the pressure would increase. Here, the water will remain in the liquid state until some external condition changes, like if the pressure is released.

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Key Concepts

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

Energy Transfer
Energy transfer refers to the movement of energy from one place or object to another. In this example involving the electrical coil and water, energy is transferred from the coil into the water, causing it to heat up. The coil provides a consistent power output of 3.50 kW, which is crucial for calculating how much energy is being transferred into the water over time. Understanding energy transfer is vital. Here, 97% of the energy from the coil successfully heats the water, showing efficient energy usage. The remaining 3%= is not lost but dissipated in other forms like heat radiating into the air, sound, or minimal mechanical vibrations, especially in a well-insulated system. These losses are typical in real-world energy applications, making 100% efficient energy transfer nearly impossible. Key takeaways from this concept include:
  • Energy transfer is the core of changing the water鈥檚 temperature here
  • Not all energy transfer results in useful work, as shown by the 3% energy loss
  • Monitoring and adjusting energy transfer can improve system efficiency
Specific Heat Capacity
Specific heat capacity is a material-specific constant that tells how much energy is necessary to raise the temperature of a unit mass by one degree Celsius. In this case, water has a specific heat capacity of approximately 4.18 J/g掳C. Understanding specific heat capacity helps in predicting how the temperature of a given substance will change as energy is added. For water, large amounts of energy are needed to raise its temperature due to its relatively high specific heat capacity.This property's significance in our example:
  • With a mass of 20 kg, calculating energy needs becomes manageable using specific heat capacity
  • Equates incoming energy to predictable temperature change: \[\Delta T = \frac{\dot{W} \cdot f \cdot dt}{m \cdot c}\]
  • For the given scenario, how quickly water reaches the desired temperature highly depends on the specific heat capacity
Understanding and using this property helps in crafting heating systems effectively and efficiently managing energy resources.
Boiling Point
The boiling point is the temperature at which a liquid turns into vapor. Water usually boils at 100掳C at 1 atm pressure. In our exercise, the water heated by the coil aims to reach this boiling point. However, achieving it doesn鈥檛 necessarily mean boiling will occur. Since the water is in a closed vessel, reaching its boiling point won鈥檛 turn it to steam unless the atmospheric conditions, mainly pressure, allow it. It's essential to evaluate:
  • The boiling point is not just a temperature target but dependent on environmental factors
  • In a closed vessel, increasing pressure hinders boiling, even at 100掳C
  • Creating steam requires reducing excess pressure or changing vessel position to allow vaporization
Thus, understanding the boiling point involves both a grasp of temperature requirements and contextual pressure dynamics.

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Most popular questions from this chapter

A kettle containing 3.00 liters of water at a temperature of \(18^{\circ} \mathrm{C}\) is placed on an electric stove and begins to boil in three minutes. (a) Write an energy balance on the water and determine an expression for \(d T / d t,\) neglecting evaporation of water before the boiling point is reached, and provide an initial condition. Sketch a plot of \(T\) versus \(t\) from \(t=0\) to \(t=4\) minutes. (b) Calculate the average rate (W) at which heat is being added to the water. Then calculate the rate (g/s) at which water vaporizes once boiling begins. (c) The rate of heat output from the stove element differs significantly from the heating rate calculated in Part (b). In which direction, and why?

The following chemical reactions take place in a liquid-phase batch reactor of constant volume \(V\). $$\begin{aligned} &\mathrm{A} \rightarrow 2 \mathrm{B} \quad r_{1}[\mathrm{mol} \mathrm{A} \text { consumed } /(\mathrm{L} \cdot \mathrm{s})]=0.100 C_{\mathrm{A}}\\\ &\mathbf{B} \rightarrow \mathbf{C} \quad r_{2}[\mathrm{mol} \mathbf{C} \text { generated } /(\mathbf{L} \cdot \mathbf{s})]=0.200 C_{\mathrm{B}}^{2} \end{aligned}$$ where the concentrations \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) are in mol/L. The reactor is initially charged with pure \(\mathrm{A}\) at a concentration of 1.00 mol/L. (a) Write expressions for ( \(i\) ) the rate of generation of \(\mathrm{B}\) in the first reaction and (ii) the rate of consumption of \(\mathrm{B}\) in the second reaction. (If this takes you more than about 10 seconds, you're missing the point.) (b) Write mole balances on A, B, and C, convert them into expressions for \(d C_{\mathrm{A}} / d t, d C_{\mathrm{B}} / d t\), and \(d C_{\mathrm{C}} / d t,\) and provide boundary conditions. (c) Without doing any calculations, sketch on a single graph the plots you would expect to obtain of \(C_{\mathrm{A}}\) versus \(t, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t .\) Clearly show the function values at \(t=0\) and \(t \rightarrow \infty\) and the curvature (concave up, concave down, or linear) in the vicinity of \(t=0 .\) Briefly explain your reasoning. (d) Solve the equations derived in Part (b) using a differential equation- solving program. On a single graph, show plots of \(C_{\mathrm{A} \text { versust }}, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t\) from \(t=0\) to \(t=50\) s. Verify that your predictions in Part (c) were correct. If they were not, change them and revise your explanation.

A radioactive isotope decays at a rate proportional to its concentration. If the concentration of an isotope is \(C(\mathrm{mg} / \mathrm{L}),\) then its rate of decay may be expressed as $$r_{\mathrm{d}}[\mathrm{mg} /(\mathrm{L} \cdot \mathrm{s})]=k C$$ where \(k\) is a constant. (a) A volume \(V(\mathrm{L})\) of a solution of a radioisotope whose concentration is \(C_{0}(\mathrm{mg} / \mathrm{L})\) is placed in a closed vessel. Write a balance on the isotope in the vessel and integrate it to prove that the half-life \(t_{1 / 2}\) of the isotope \(-\) by definition, the time required for the isotope concentration to decrease to half of its initial value- equals ( \(\ln 2\) )/ \(k\). (b) The half-life of \(^{56} \mathrm{Mn}\) is \(2.6 \mathrm{h}\). A batch of this isotope that was used in a radiotracing experiment has been collected in a holding tank. The radiation safety officer declares that the activity (which is proportional to the isotope concentration) must decay to \(1 \%\) of its present value before the solution can be discarded. How long will this take?

A steam coil is immersed in a stirred tank. Saturated steam at 7.50 bar condenses within the coil, and the condensate emerges at its saturation temperature. A solvent with a heat capacity of \(2.30 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\cdot} \mathrm{C}\right)\) is fed to the tank at a steady rate of \(12.0 \mathrm{kg} / \mathrm{min}\) and a temperature of \(25^{\circ} \mathrm{C},\) and the heated solvent is discharged at the same flow rate. The tank is initially filled with \(760 \mathrm{kg}\) of solvent at \(25^{\circ} \mathrm{C},\) at which point the flows of both steam and solvent are commenced. The rate at which heat is transferred from the steam coil to the solvent is given by the expression $$\dot{Q}=U A\left(T_{\mathrm{steam}}-T\right)$$ where \(U A\) (the product of a heat transfer coefficient and the coil surface area through which the heat is transferred) equals \(11.5 \mathrm{kJ} /\left(\min \cdot^{\circ} \mathrm{C}\right) .\) The tank is well stirred, so that the temperature of the contents is spatially uniform and equals the outlet temperature. (a) Prove that an energy balance on the tank contents reduces to the equation given below and supply an initial condition. \frac{d T}{d t}=1.50^{\circ} \mathrm{C} / \mathrm{min}-0.0224 T (b) Without integrating the equation, calculate the steady-state value of \(T\) and sketch the expected plot of \(T\) versus \(t,\) labeling the values of \(T_{\mathrm{b}}\) at \(t=0\) and \(t \rightarrow \infty\) (c) Integrate the balance equation to obtain an expression for \(T(t)\) and calculate the solvent temperature after 40 minutes. (d) The tank is shut down for routine maintenance, and a technician notices that a thin mineral scale has formed on the outside of the steam coil. The coil is treated with a mild acid that removes the scale and reinstalled in the tank. The process described above is run again with the same steam conditions, solvent flow rate, and mass of solvent charged to the tank, and the temperature after 40 minutes is \(55^{\circ} \mathrm{C}\) instead of the value calculated in Part (c). One of the system variables listed in the problem statement must have changed as a result of the change in the stirrer. Which variable would you guess it to be, and by what percentage of its initial value did it change?

A stirred tank contains \(1500 \mathrm{lb}_{\mathrm{m}}\) of pure water at \(70^{\circ} \mathrm{F}\). At time \(t=0,\) two streams begin to flow into the tank and one is withdrawn. One input stream is a \(20.0 \mathrm{wt} \%\) aqueous solution of \(\mathrm{NaCl}\) at \(85^{\circ} \mathrm{F}\) flowing at a rate of \(15 \mathrm{lb}_{\mathrm{m}} / \mathrm{min},\) and the other is pure water at \(70^{\circ} \mathrm{F}\) flowing at \(10 \mathrm{lb}_{\mathrm{m}} / \mathrm{min} .\) The mass of liquid in the tank is held constant at \(1500 \mathrm{lb}_{\mathrm{m}}\). Perfect mixing in the tank may be assumed, so that the outlet stream has the same \(\mathrm{NaCl}\) mass fraction \((x)\) and temperature \((T)\) as the tank contents. Also assume that the heat of mixing is zero and the heat capacity of all fluids is \(C_{p}=1 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) (a) Write differential material and energy balances and use them to derive expressions for \(d x / d t\) and \(d T / d t\) (b) Without solving the equations derived in Part (a), sketch plots of \(T\) and \(x\) as a function of time \((t)\) Clearly identify values at time zero and as \(t \rightarrow \infty\)
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