Initial and final temperatures

Now if the chemical reaction had been allowed to proceed without the performance of any external electrical work, say in a calorimeter, so that the initial and final temperatures of the system are both T, the change of intrinsic energy would have been the same as that occurring in the process described above, as we know from the First Law. Thus the heat of reaction, Q will be equal to the increase of intrinsic energy ... 

In the case of a storage tank with liquor of mass m and specific heat C heated by steam condensing in a helical coil, it may be assumed that the overall transfer coefficient U is constant. If 7 is the temperature of the condensing steam, Tt and To the initial and final temperatures of the liquor, and A the area of heat transfer surface, and T k the temperature of the liquor at any time t, then the rate of transfer of heat is given by ... 

Initial and final temperature and feed rate are taken as parameters to be optimized, whereby the other variables are optimized in the same run. Temperature and feed rate between these two points are assumed to be straight lines connecting the initial and final values. The optimal values of variables obtained in the first step are taken as initial guesses for optimization. 

This is the same problem as Example I8.6. In that problem the temperature change was specified. In this example, the initial and final temperatures arc given, but the temperature change is the same 17.0°C. The answer is again 7.11 kJ. 

Ans. This is the same as Problem 18.10 except that the initial and final temperatures are given instead of the temperature change. 

Since we have provided initial and final temperatures but have not specified any reactants, the program traces a polythermal path for a closed system (see Chapter 14). The fluid s pH (Fig. 23.1) changes with temperature from its initial value of 5 at 250 °C to less than 4 at 25 °C. The change is entirely due to variation in the stabilities of the aqueous species in solution. As shown in Figure 23.2, the H+ concentration increases in response to the dissociation of the HC1 ion pair,... 

Be familiar with the basic measurements required for the experiments. For example, in a calorimetry experiment you do not measure the change in temperature, you calculate it. You measure the initial and final temperatures. 

Mass and possible volume measurements, along with the initial and final temperatures, are needed. Remember you measure the initial and final temperature so you can calculate the change in temperature. 

You calculate the change in temperature from your measured initial and final temperatures. You do not need to give a lot of detail when listing the required measurements, but you need to be very specific in what you measure. Many students, have lost exam points for not clearly distinguishing between measured and calculated values. 

If, as shown in figure 12.9a, the whole transformation strictly occurs in the interval (Te—(), IK —> 7 c, then the corresponding enthalpy change// can be calculated from equation 12.37 by using the areas of two peaks only (i) the large peak recorded in the main experiment between Te—0.1K and Te, and (ii) the corresponding peak for the zero line. In this case, the process is quasi-isothermic because the initial and final temperatures differ by only 0.1 K. Therefore, unless... 

For the reversible adiabatic expansion, we can see from Equation (5.42) that the final temperature T2 must be less than Ti, because W is negative and Cy is always positive. Thus, the adiabatic reversible expansion is accompanied by a temperature drop, and W, AU, and AH can be calculated from the measured initial and final temperatures using Equations (5.42) and (5.43). 

For the reversible adiabatic expansion, a definite expression can be derived to relate the initial and final temperatures to the respective volumes or pressures if we assume that the heat capacity is independent of temperature. This assumption is exact at all temperatures for monatomic gases and above room temperature for diatomic gases. Again we start with Equation (5.39). Recognizing the restriction of reversibility, we obtain... 

For crystallization by acid addition, the effects of acid addition rate, agitation, and initial solution composition were examined. The experiments had constant initial and final temperatures, although there were some variations in temperature during a run because of the heat of crystallization of L-isoleucine hydrochloride (L-Ile HCl H20) and heat of solution of acid. The acid used was in the form of 37% HCl. The initial conditions of the batch corresponded to an HCl concentration giving the maximum solubility of L-Ile (1). A schematic of the batch crystallizer is shown in Figure 3. 

Clearly, this format takes more time to write, but consider the benefit to the student challenged to calculate an enthalpy change resulting from an isobaric and isoplethic heating. The expression itself tells the student what must be done, the initial and final temperatures must be known and an expression for a change in enthalpy depending only on temperature must be found because pressure and composition are constant. What does AH tell the student Push this example further. 

Initial and final temperatures. The temperature range used for a separation in temperature-programmed chromatography. 

Use the thermometer diagrams to record all of the initial and final temperatures in the table. 

This is not the final temperature (choice A is incorrect). It is the temperature difference between the initial and final temperature. 

Unlike work and heat, the property changes of the system for step d can be computed, since they depend solely on the initial and final states, and these are known. The internal energy and enthalpy of an ideal gas are functions of temperature only. Therefore, A Ud and Atfd are zero, because the initial and final temperatures are both 27°C. The first law applies to irreversible as well as to reversible processes, and for step d it becomes... 

III. Adiabatic Compression.—It is a matter of common knowledge that when a gas is rapidly compressed, a rise in temperature takes place, the relationship between the initial and final temperatures being given by the expression... 

You know the mass of the water. You also know the initial and final temperatures of the water. 

You have enough information to solve this problem using Q = me AT. Use the initial and final temperatures to calculate AT. You need the specific heat capacity (c) of liquid water. This is given in Table 14.2 (4.184 J/g-°C). Because you are only concerned with the water, you will not use the mass of the ice. 

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