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A thermostat control with dial markings from 0 to 100 is used to regulate the temperature of an oil bath. A calibration plot on logarithmic coordinates of the temperature, \(T(\text { ( } \mathrm{F} \text { ), versus the dial setting, } R\), is a straight line that passes through the points \(\left(R_{1}=20.0, T_{1}=110.0^{\circ} \mathrm{F}\right)\) and \(\left(R_{2}=40.0, T_{2}=250.0^{\circ} \mathrm{F}.\right)\) (a) Derive an equation for \(T\left(^{\circ} \mathrm{F}\right)\) in terms of \(R\). (b) Estimate the thermostat setting needed to obtain a temperature of \(320^{\circ} \mathrm{F}\). (c) Suppose you set the thermostat to the value of \(R\) calculated in Part (b), and the reading of athermocouple mounted in the bath equilibrates at \(295^{\circ} \mathrm{F}\) instead of \(320^{\circ} \mathrm{F}\). Suggest several possible explanations.

Short Answer

Expert verified
The function that relates \(R\) and \(T\) on a logarithmic plot can be expressed as \(\log(T) = aR + b\). Substituting \(320^{\circ} F\) and solving the equation will provide the thermostat setting required. Possible reasons for a reading discrepancy could include calibration errors, uneven distribution of heat in the oil bath or errors with the thermocouple.

Step by step solution

01

Deriving the equation

Given two points \((R_1, T_1) = (20.0, 110.0^{\circ} F)\) and \((R_2, T_2) = (40.0, 250.0^{\circ} F)\), as well as the known fact that this is a straight line on a logarithmic scale, we can conclude that this is a logarithmic function of the form \( y = a\log(x) + b\). We can convert this to exponential form to find the constants a and b, giving us the equation \(T = 10 ^ {aR + b}\). This can be changed into a linear equation by applying a logarithm to both sides, resulting in \(\log(T) = aR + b\). By inputting the known variables, we obtain two equations in the form: \(\log(T1) = aR_1 + b\) and \(\log(T2) = aR_2 + b\). We can solve this system to obtain values for a and b.
02

Estimating the Thermostat Setting

Given the equation found in Step 1, and the desire to attain a temperature of \(320^{\circ} F\), we substitute \(\log(T)\) for \(\log(320)\) in the equation, and solve for \(R\).
03

Evaluating Thermostat Accuracy

If the measured result is different from the expected, there could be a few explanations. One could be a calibration error in the thermostat, perhaps due to wear and tear or a lack of proper maintenance and regular calibration. It could also be that the temperature in the oil bath was not properly mixed, leading to a non-uniform temperature reading. Lastly, it could be an error in the thermocouple, such as an incorrect reading or improper placement within the bath.

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Key Concepts

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

Logarithmic Temperature Control
In many industrial and laboratory settings, maintaining temperature at a precise level is crucial for consistent results. Logarithmic temperature control is an advanced method that offers a solution for regulating temperature with high accuracy. By using a logarithmic scale instead of a linear one, the control system provides more sensitivity at lower temperatures where more precise adjustments are often needed. This type of control is particularly useful when a process requires a very gradual increase or decrease in temperature, as it can provide finer tuning over a wide range of temperatures.

When it comes to thermostat calibration on a logarithmic scale, as seen in the exercise, the calibration plot is a straight line on a log graph. Here, the relationship between the dial setting, typically represented by the variable 'R', and the temperature, 'T', is expressed by an equation that features logarithmic components. Deriving this equation involves determining constants that align with known points on the calibration curve, allowing for accurate temperature control based on the thermostat's dial setting. An understanding of logarithmic functions is essential in developing the calibration equation and ensuring that the thermostat provides the correct output for the desired temperature setting.
Calibration Plot Analysis
Calibration plot analysis plays a pivotal role in ensuring the accuracy of temperature measurements and control devices. It serves as a visual representation of the relationship between an instrument's reading and the actual temperature. For a thermostat controlling an oil bath, as outlined in the exercise, a calibration plot is created by mapping known temperature readings against their corresponding dial settings. Since the plot is expected to be a straight line on logarithmic coordinates, the slope and intercept of this line can be used to derive a precise calibration equation.

In this specific scenario, the given points \(\left(R_{1}=20.0, T_{1}=110.0^{\circ} \mathrm{F}\right)\) and \(\left(R_{2}=40.0, T_{2}=250.0^{\circ} \mathrm{F}\right)\) are used to establish two equations, which when solved, yield the constants required for the temperature control formula. Conducting a proper calibration plot analysis ensures the accuracy of the derived equation, which in turn allows for reliable temperature setting and control in practical applications.
Thermocouple Temperature Measurement
Thermocouples are a type of temperature sensor commonly used to measure and control temperature in various processes. They work based on the Seebeck effect, which occurs when two dissimilar metals are joined at one end and produce a voltage proportional to the temperature difference between the two ends. Thermocouple temperature measurement is highly valued for its precision, simplicity, and ability to measure a wide range of temperatures.

In the context of the exercise, a thermocouple is used to verify the actual temperature of the oil bath, which should correspond to the temperature indicated by the thermostat setting. If there is a discrepancy, as evidenced by the thermocouple reading equilibrating at \(295^{\circ} \mathrm{F}\) instead of the expected \(320^{\circ} \mathrm{F}\), it signals a potential issue. Possible explanations for the difference might include calibration errors in the thermostat or the thermocouple, environmental factors affecting the measurement, or even errors in the setup, such as improper placement of the thermocouple in the oil bath. Understanding how thermocouples function and potential sources of error can help in troubleshooting such discrepancies and maintaining the fidelity of temperature-dependent processes.

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Most popular questions from this chapter

Drop-on-demand (DoD) technology is an emerging form of drug delivery in which a reservoir is filled with a solution of an active pharmaceutical ingredient (API) dissolved in a volatile liquid, and a device sprays nanometer-scale drops of the solution onto an edible substrate, such as a small strip the size of a stick of chewing gum. The liquid evaporates very rapidly, causing the API to crystallize on the substrate. The exact dose required by a patient can be administered based on the known concentration of the API in the reservoir and the volume of solution deposited on the sa1) For a man of \(245 \mathrm{lb}_{m}\) dosage is \(2.2 \mathrm{mL}=5.3 \cdot 10^{20}\) drops a2) For a child of \(65 ~ l b_{m}\) dosage is \(0.60 \mathrm{mL}=1.4 \cdot 10^{20}\) drops b) \(V_{\text {dose}}=\frac{D \cdot W_{p} \cdot 0.4536 \mathrm{kg} / \mathrm{lb}_{f}}{M W_{A P I} \cdot m_{s}} \cdot \frac{1000 \mathrm{mL}}{1 \mathrm{L}}\) c) \(\frac{A}{V}=\frac{3}{\pi},\) the nanometric drops have larger area, through which the water can evaporate faster.ubstrate, enabling greater dosage accuracy than can be provided by administering fractions of tablets. (a) A DoD device is charged with a 1.20 molar solution of ibuprofen (the API) in \(n\) -hexane. The molecular weight of ibuprofen is \(206.3 \mathrm{g} / \mathrm{mol} .\) If a prescribed dosage is \(5.0 \mathrm{mg}\) ibuprofen/kg patient weight, how many milliliters of solution should be sprayed for a 245 -pound man and a 65 -pound child? How many drops are in each dose, assuming that each drop is a sphere with a radius of \(1 \mathrm{nm} ?\) (b) The DoD device is to be automated, so that the operator enters a patient's body weight into a computer that determines the required solution volume and causes that volume to be sprayed on the substrate. Derive a formula for the volume, \(V_{\text {dare }}(\mathrm{mL}),\) in terms of the following variables: \(M_{\mathrm{s}}(\operatorname{mol} \mathrm{API} / \mathrm{L})=\) molarity of reservoir solution \(\mathrm{SG}_{\mathrm{s}}=\) specific gravity of reservoir solution \(\mathrm{MW}_{\mathrm{API}}(\mathrm{g} / \mathrm{mol})=\) molecular weight of \(\mathrm{API}\) \(D(\mathrm{mg} \mathrm{APl} / \mathrm{kg} \text { body weight })=\) prescribed dosage \(W_{\mathrm{p}}\left(\mathrm{b}_{\mathrm{f}}\right)=\) patient's weight Check your formula by verifying your solution to Part (a). (c) Calculate the surface-to-volume ratio of a sphere of radius \(r .\) Then calculate the total drop surface area of \(1 \mathrm{mL}\left(=1 \mathrm{cm}^{3}\right)\) of the solution if it were sprayed as drops of (i) radius \(1 \mathrm{nm}\) and (ii) \(1 \mathrm{mm}\). Speculate on the likely reason for spraying nanoscale drops instead of much larger drops.

Aqueous solutions of the amino acid \(\mathrm{L}\) -isoleucine (Ile) are prepared by putting 100.0 grams of pure water into each of six flasks and adding different precisely weighed quantities of Ile to each flask. The densities of the solutions at \(50.0 \pm 0.05^{\circ} \mathrm{C}\) are then measured with a precision densitometer, with the following results: $$\begin{array}{|l|l|l|l|l|l|l|}\hline r\left(\mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\right) & 0.0000 & 0.8821 & 1.7683 & 2.6412 & 3.4093 & 4.2064 \\\\\hline \rho\left(\mathrm{g} \text { solution } / \mathrm{cm}^{3}\right) & 0.98803 & 0.98984 & 0.99148 & 0.99297 & 0.99439 & 0.99580 \\\\\hline\end{array}$$ (a) Plot a calibration curve showing the mass ratio, \(r,\) as a function of solution density, \(\rho,\) and fit a straight line to the data to obtain an equation of the form \(r=a \rho+b\) (b) The volumetric flow rate of an aqueous Ile solution at a temperature of \(50^{\circ} \mathrm{C}\) is \(150 \mathrm{L} / \mathrm{h}\). The density of a sample of the stream is measured at \(50^{\circ} \mathrm{C}\) and found to be \(0.9940 \mathrm{g} / \mathrm{cm}^{3} .\) Use the calibration equation to estimate the mass flow rate of Ile in the stream (kg IIe/h). (c) It has just been discovered that the thermocouple used to measure the stream temperature was poorly calibrated and the temperature was actually \(47^{\circ} \mathrm{C}\). Would the Ile mass flow rate calculated in Part \((\mathrm{b})\) be too high or too low? State any assumption you make and briefly explain your reasoning.

Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, \(\left.\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right),\) has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6 .) In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite \(\left(\mathrm{CaCO}_{3}\right)\) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous \(\mathrm{CaSO}_{4}\) solution. The slurry flows from the crystallizer to a filter in which the particles are collected as a filter cake. The filter cake, which is 95.0 wiff solid gypsum and the remainder CaSO_solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO \(_{4}\) as product. A flowchart and relevant process data are given below. Solids content of slurry leaving crystallizer: \(0.35 \mathrm{kg} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) slurry \(\mathrm{CaSO}_{4}\) content of slurry liquid: \(0.209 \mathrm{g} \mathrm{CaSO}_{4} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) Specific gravities: \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), 2.32 ;\) liquid solutions, 1.05 (a) Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer). (b) Takea basis of one liter of solution leaving the crystallizer and calculate the mass (kg) and volume (L) of solid gypsum, the mass of \(\mathrm{CaSO}_{4}\) in the gypsum, and the mass of \(\mathrm{CaSO}_{4}\) in the liquid solution. (c) Calculate the percentage recovery of \(\mathrm{CaSO}_{4}-\) that is, the percentage of the total \(\mathrm{CaSO}_{4}\) (precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous \(\mathrm{CaSO}_{4}\) (d) List five potential negative consequences of discharging \(\mathrm{H}_{2} \mathrm{SO}_{4}\) into the river passing the plant.

An open-end mercury manometer is to be used to measure the pressure in an apparatus containing a vapor that reacts with mercury. A 10 -cm layer of silicon oil \((\mathrm{SG}=0.92)\) is placed on top of the mercury in the arm attached to the apparatus. Atmospheric pressure is \(765\) \(\mathrm{mm}\) Hg. (a) If the level of mercury in the open end is 365 mm below the mercury level in the other arm, what is the pressure (mm Hg) in the apparatus? (b) When the instrumentation specialist was deciding on a liquid to put in the manometer, she listed several properties the fluid should have and eventually selected silicon oil. What might the listed properties have been?

An inclined manometer is a useful device for measuring small pressure differences. The formula given in Section 3.4 for the pressure difference in terms of the liquid-level difference \(h\) remains valid, but while \(h\) would be small and difficult to read for a small pressure drop if the manometer were vertical, \(L\) can be made quite large for the same pressure drop by making the angle of the inclination, \(\theta,\) small. (a) Derive a formula for \(h\) in terms of \(L\) and \(\theta\) (b) Suppose the manometer fluid is water, the process fluid is a gas, the inclination of the manometer is \(\theta=15^{\circ},\) and a reading \(L=8.7 \mathrm{cm}\) is obtained. What is the pressure difference between points? and?? (c) The formula you derived in Part (a) would not work if the process fluid were a liquid instead of a gas. Give one definite reason and another possible reason.

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