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Chapter 10: Problem 26

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?

Short Answer

Expert verified
The energy balance for the water kettle system is \(q_{in} = m c \frac{dT}{dt}\) and the initial condition is \(T(0) = 18^{\circ}C\). The average rate at which heat is being added to the water is determined by the energy required to heat the water to boil and the time it takes to boil. The rate of water vaporization once boiling begins equates the heat addition rate to the heat required for vaporization. The actual heat output from the stove element is greater than the heating rate calculated due to heat losses.

Step by step solution

01

Energy Balance

First an energy balance for the water kettle system is written. It is given by \[q_{in} - q_{out} = m c \frac{dT}{dt}\] where \(q_{in}\) is the energy input into the system, \(q_{out}\) is the energy leaving the system (which is zero since no evaporation is to be considered before boiling point), \(m\) is the mass of the water, \(c\) is the specific heat capacity of water, and \(\frac{dT}{dt}\) is the rate of change of temperature with respect to time. Given that \[m = \text{volume of water} \times \text{density of water} = 3.0 \text{L} \times 1 \text{kg/L} = 3.0 \text{kg}\] and \(c = 4.186 \times 10^3 \text{J/(kg掳C)}\), the equation simplifies to \[q_{in} = m c \frac{dT}{dt}\] This gives the initial condition of \[T(0) = 18^{\circ}\text{C}\] Furthermore, a plot of \(T\) vs \(t\) would be a straight line from \(18^{\circ}\text{C}\) to \(\approx 100^{\circ}\text{C}\) between \(t=0\) to \(t=3\) minutes since heat up happens at a constant rate and then remain constant at \(\approx 100^{\circ}\text{C}\) for \(t>3\) minutes as boiling starts at \(100^{\circ}\text{C}\).
02

Calculate the Average Heating Rate

First determine the energy required to heat the water to boil. This is given by \[q_{in} = m c (T_{final} - T_{initial})\] Where \(T_{final} = 100^{\circ}\text{C}\) (boiling point of water) and \(T_{initial} = 18^{\circ}\text{C}\). The energy \(q_{in}\) is then divided by the time to boil (3 minutes or 180 seconds) to get the average rate of heat addition \[P_{avg} = \frac{q_{in}}{t}\]. The rate of water vaporization once boiling begins can be calculated by equating the heat addition rate to the heat required for vaporization, \(m_{vap} h_{vap} / t\), where \(m_{vap}\) is the rate of water vaporization and \(h_{vap}\) is the latent heat of vaporization for water.
03

Comparing Heat Output Rates

The calculations in part (b) do not account for certain heat losses to the surroundings and the kettle itself. Therefore, the actual heat output from the stove element would be greater than the heating rate calculated. This ensures that the kettle receives enough heat to maintain the boiling process despite these losses.

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity is a measure of how much energy is needed to raise the temperature of a substance by one degree Celsius. For water, this value is quite high at 4,186 J/(kg掳C). This means it takes 4,186 joules of energy to heat one kilogram of water by just one degree Celsius. This is why water is excellent at storing and transferring energy.
In the context of our kettle exercise, the specific heat capacity is crucial as it helps us understand the energy needed to heat 3kg of water from its initial temperature of 18掳C up to the boiling point of 100掳C. Since you're heating a significant amount of water, the high specific heat capacity lets you calculate the energy required, setting the foundation for understanding the energy transfer in the system.
Latent Heat of Vaporization
Once water reaches its boiling point, additional energy is not used for further increasing the temperature but for phase change, turning liquid water into vapor. This energy is known as the latent heat of vaporization. For water, this value is 2260 kJ/kg.
During the boiling process, water absorbs this energy without any rise in temperature, allowing it to change phases. In the exercise, once the water in the kettle reaches 100掳C, any heat provided contributes to the process of vaporization rather than increasing temperature. Knowing this helps differentiate between the heating phase and the vaporization phase in energy calculations, making it easier to compute the rate of vaporization.
Temperature Change
Temperature change refers to the difference between the initial and final temperatures of a system. In this exercise, the water undergoes a temperature change from 18掳C to 100掳C. This change is driven by energy absorbed from the stove.
Using the formula \( q = mc\Delta T \), where \( q \) is the heat added, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change, you can calculate the total energy required to heat up the water. This calculation also helps to understand how the energy in the system is transferred efficiently and why certain amounts of heat result in specific temperature changes. This helps provide insights into thermal dynamics and energy interactions.
Heat Loss
Heat loss is an important factor to consider because in real-world scenarios, not all the heat produced by the stove ends up in the water; some is lost to the surrounding environment. This can occur through convection, conduction, or radiation from the kettle to the air, and is why the actual energy output from the stove is higher than the energy calculated based solely on the water's heating needs.
By acknowledging heat loss, you gain a better understanding of how energy does not perfectly convert into work or heat transfer. It's crucial for calculating the stove's total heat output, as you need to consider additional energy to compensate for the loss to maintain boiling. This concept helps illustrate inefficiencies in energy systems and why real-life scenarios often require more energy than theoretical calculations suggest.

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Most popular questions from this chapter

A tracer is used to characterize the degree of mixing in a continuous stirred tank. Water enters and leaves the mixer at a rate of \(\dot{V}\left(\mathrm{m}^{3} / \mathrm{min}\right) .\) Scale has built up on the inside walls of the tank, so that the effective volume \(V\left(\mathrm{m}^{3}\right)\) of the tank is unknown. At time \(t=0,\) a mass \(m_{0}(\mathrm{kg})\) of the tracer is injected into the tank and the tracer concentration in the outlet stream, \(C\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) is monitored. (a) Write a differential balance on the tracer in the tank in terms of \(V, C,\) and \(\dot{V},\) assuming that the tank contents are perfectly mixed, and convert the balance into an equation for \(d C / d t\). Provide an initial condition, assuming that the injection is rapid enough so that all of the tracer may be considered to be in the tank at \(t=0 .\) Without doing any calculations, sketch a plot of \(C\) versus \(t\) labeling the value of \(C\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\) (b) Integrate the balance to prove that $$C(t)=\left(m_{0} / V\right) \exp (-\dot{V} t / V)$$ (c) Suppose the flow rate through the mixer is \(\dot{V}=30.0 \mathrm{m}^{3} / \mathrm{min}\) and that the following data are taken: (For example, at \(t=1\) min, \(C=0.223 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{3}\).) Verify graphically that the tank is functioning as a perfect mixer- -that is, that the expression of Part (b) fits the data- -and determine the effective volume \(V\left(\mathrm{m}^{3}\right)\) from the slope of your plot. (d) A solution of a radioactive element with a fairly short half-life (see Problem 10.16 ) is often used as a tracer for applications like the one in this problem. The advantage of doing so is that the concentration of the tracer at the outlet can be measured with a sensitive radiation detector mounted outside the exit pipe rather than having to draw fluid samples from the pipe and analyze them. What is a potential drawback of radiotracers? Why is it important that the half-life of the tracer be neither too short nor too long?

Methane is generated via the anaerobic decomposition (biological degradation in the absence of oxygen) of solid waste in landfills. Collecting the methane for use as a fuel rather than allowing it to disperse into the atmosphere provides a useful supplement to natural gas as an energy source. If a batch of waste with mass \(M\) (tonnes) is deposited in a landfill at \(t=0,\) the rate of methane generation at time \(t\) is given by $$\dot{V}_{\mathrm{CH}_{4}}(t)=k L_{0} M_{\text {waste }} e^{-k t}$$ where \(\dot{V}_{\mathrm{CH}_{4}}\) is the rate at which methane is generated in standard cubic meters per year, \(k\) is a rate constant, \(L_{0}\) is the total potential yield of landfill gas in standard cubic meters per tonne of waste, and \(M_{\text {watte is the tonnes of waste in the landfill at } t=0}\). (a) Starting with Equation 1, derive an expression for the mass generation rate of methane, \(\dot{M}_{\mathrm{CH}_{4}}(t)\) Without doing any calculations, sketch the shape of a plot of \(M_{\mathrm{CH}, \text { versus } t \text { from } t=0 \text { to } t=3 \mathrm{y},}\) and graphically show on the plot the total masses of methane generated in Years \(1,2,\) and \(3 .\) Then derive an expression for \(M_{\mathrm{CH}_{4}}(t),\) the total mass of methane (tonnes) generated from \(t=0\) to a time \(t\) (b) A new landfill has a yield potential \(L_{0}=100\) SCM CH \(_{4}\) /tonne waste and a rate constant \(k=0.04 \mathrm{y}^{-1} .\) At the beginning of the first year, 48,000 tonnes of waste are deposited in the landfill. Calculate the tonnes of methane generated from this deposit over a three-year period. (c) A colleague solving the problem of Part (b) calculates the methane produced in three years from the \(4.8 \times 10^{4}\) tonnes of waste as $$M_{\mathrm{CH}_{4}}(t=3)=\dot{M}_{\mathrm{CH}_{4}}(t=0) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=1) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=2) \times 1 \mathrm{y}$$ where \(\dot{M}_{\mathrm{CH}_{4}}\) is the first expression derived in Part (a). Briefly state what has been assumed about the rate of methane generation. Calculate the value determined with this method and the percentage error in the calculation. Show graphically what the calculated value corresponds to on another sketch of \(M_{\mathrm{CH}_{4}}\) versus \(t\) (d) The following amounts of waste are deposited in the landfill on January 1 in each of three consecutive years. Exploratory Exercises - Research and Discover (e) Explain in your own words the benefits of reducing the release of methane from landfills and of using the methane as a fuel instead of natural gas. (f) One way to avoid the environmental hazard of methane generation is to incinerate the waste before it has a chance to decompose. What problems might this alternative process introduce?

An immersed electrical heater is used to raise the temperature of a liquid from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20.0 min. The combined mass of the liquid and the container is \(250 \mathrm{kg}\), and the mean heat capacity of the system is 4.00 kJ/(kg.'C). The liquid decomposes explosively at 85"C. At 10: 00 a.m. a batch of liquid is poured into the vessel, and the operator turns on the heater and answers a call on his cell phone. Ten minutes later, his supervisor walks by and looks at the computer display of the power input. This what she sees. The supervisor immediately shuts off the heater and charges off to pass on to the operator several brief observations that come to her mind. (a) Calculate the required constant power input \(\dot{Q}(\mathrm{k} \mathrm{W})\), neglecting energy losses from the container. (b) Write and integrate using Simpson's rule (Appendix A.3) an energy balance on the system to estimate the system temperature at the moment the heater is shut off. Use the following data from the recorder chart: $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|} \hline t(\mathrm{s}) & 0 & 30 & 60 & 90 & 120 & 150 & 180 & 210 & 240 & 270 & 300 \\ \hline \dot{Q}(\mathrm{kW}) & 33 & 33 & 34 & 35 & 37 & 39 & 41 & 44 & 47 & 50 & 54 \\ \hline \end{array}$$ $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline t(\mathrm{s}) & 330 & 360 & 390 & 420 & 450 & 480 & 510 & 540 & 570 & 600 \\ \hline \dot{Q}(\mathrm{kW}) & 58 & 62 & 66 & 70 & 75 & 80 & 85 & 90 & 95 & 100 \\\ \hline \end{array}$$ (c) Suppose that if the heat had not been shut off, \(\dot{Q}\) would have continued to increase linearly at a rate of \(10 \mathrm{kW} / \mathrm{min}\). At what time would everyone in the plant realize that something was wrong?

A gas that contains \(\mathrm{CO}_{2}\) is contacted with liquid water in an agitated batch absorber. The equilibrium solubility of \(\mathrm{CO}_{2}\) in water is given by Henry's law (Section \(6.4 \mathrm{b}\) ) $$C_{\mathrm{A}}=p_{\mathrm{A}} / H_{\mathrm{A}}$$ where \(C_{\mathrm{A}}\left(\mathrm{mol} / \mathrm{cm}^{3}\right)=\) concentration of \(\mathrm{CO}_{2}\) in solution, \(p_{\mathrm{A}}(\mathrm{atm})=\) partial pressure of \(\mathrm{CO}_{2}\) in the gas phase, and \(H_{\mathrm{A}}\left[\mathrm{atm} /\left(\mathrm{mol} / \mathrm{cm}^{3}\right)\right]=\) Henry's law constant. The rate of absorption of \(\mathrm{CO}_{2}\) (i.e., the rate of transfer of \(\mathrm{CO}_{2}\) from the gas to the liquid per unit area of gas-liquid interface) is given by the expression $$r_{\mathrm{A}}\left[\operatorname{mol} /\left(\mathrm{cm}^{2} \cdot \mathrm{s}\right)\right]=k\left(C_{\mathrm{A}}^{*}-C_{\mathrm{A}}\right)$$ where \(C_{A}=\) actual concentration of \(\mathrm{CO}_{2}\) in the liquid, \(C_{\mathrm{A}}^{*}=\) concentration of \(\mathrm{CO}_{2}\) in the liquid that would be in equilibrium with the \(\mathrm{CO}_{2}\) in the gas phase, and \(k(\mathrm{cm} / \mathrm{s})=\) a mass transfer coefficient. The gas phase is at a total pressure \(\mathrm{P}\left(\text { atm) and contains } y_{\mathrm{A}}\left(\mathrm{mol} \mathrm{CO}_{2} / \mathrm{mol}\text { gas), and the liquid }\right.\right.\) phase initially consists of \(V\left(\mathrm{cm}^{3}\right)\) of pure water. The agitation of the liquid phase is sufficient for the composition to be considered spatially uniform, and the amount of \(\mathrm{CO}_{2}\) absorbed is low enough for \(P, V,\) and \(y_{\mathrm{A}}\) to be considered constant throughout the process. (a) Derive an expression for \(d C_{\mathrm{A}} / d t\) and provide an initial condition. Without doing any calculations, sketch a plot of \(C_{\mathrm{A}}\) versus \(t,\) labeling the value of \(C_{\mathrm{A}}\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\) Give a physical explanation for the asymptotic value of the concentration. (b) Prove that $$C_{\mathrm{A}}(t)=\frac{p_{\mathrm{A}}}{H_{\mathrm{A}}}[1-\exp (-k S t / V)]$$ where \(S\left(\mathrm{cm}^{2}\right)\) is the effective contact area between the gas and liquid phases. (c) Suppose the system pressure is 20.0 atm, the liquid volume is 5.00 liters, the tank diameter is \(10.0 \mathrm{cm},\) the gas contains 30.0 mole \(\% \mathrm{CO}_{2},\) the Henry's law constant is \(9230 \mathrm{atm} / \mathrm{mole} / \mathrm{cm}^{3}\) ), and the mass transfer coefficient is \(0.020 \mathrm{cm} / \mathrm{s}\). Calculate the time required for \(C_{\mathrm{A}}\) to reach \(0.620 \mathrm{mol} / \mathrm{L}\) if the gas-phase properties remain essentially constant. (d) If A were not \(\mathrm{CO}_{2}\) but instead a gas with a moderately high solubility in water, the expression for \(C_{\mathrm{A}}\) given in Part (b) would be incorrect. Explain where the derivation that led to it would break down.

Ninety kilograms of sodium nitrate is dissolved in \(110 \mathrm{kg}\) of water. When the dissolution is complete (at time \(t=0\) ), pure water is fed to the tank at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min}),\) and solution is withdrawn from the tank at the same rate. The tank may be considered perfectly mixed. (a) Write a total mass balance on the tank and use it to prove that the total mass of liquid in the tank remains constant at its initial value. (b) Write a balance on sodium nitrate, letting \(x(t, \dot{m})\) equal the mass fraction of \(\mathrm{NaNO}_{3}\) in the tank and outlet stream. Convert the balance into an equation for \(d x / d t\) and provide an initial condition. (c) On a single graph of \(x\) versus \(t,\) sketch the shapes of the plots you would expect to obtain for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min} .\) (Don't do any calculations.) Explain your reason- ing, using the equation of Part (b) in your explanation. (d) Separate variables and integrate the balance to obtain an expression for \(x(t, \dot{m})\). Check your solution. Then generate the plots of \(x\) versus \(t\) for \(\dot{m}=50 \mathrm{kg} / \mathrm{min}, 100 \mathrm{kg} / \mathrm{min},\) and \(200 \mathrm{kg} / \mathrm{min}\) and show them on a single graph. (A spreadsheet is a convenient tool for carrying out this step.) (e) If \(\dot{m}=100 \mathrm{kg} / \mathrm{min}\), how long will it take to flush out \(90 \%\) of the sodium nitrate originally in the tank? How long to flush out 99\%? 99.9\%? (f) The stream of water enters the tank at a point near the top, and the exit pipe from the tank is located on the opposite side toward the bottom. One day the plant technician forgot to turn on the mixing impeller in the tank. On the same chart, sketch the shapes of the plots of \(x\) versus \(t\) you would expect to see with the impeller on and off, clearly showing the differences between the two curves at small values and large values of \(t .\) Explain your reasoning.
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