In the first part of this exercise we determine the calorimeter constant, C_cal.

Question

In the first part of this exercise we determine the calorimeter constant, C_cal. We measure the temperatures of samples of cold and hot water for several minutes. At 3.0 minutes, we mix the water samples and follow the temperature of the resultant mixture for 7 minutes. The hot and cold water temperatures are extrapolated forward to the time of mixing (3 minutes). The temperature of the mixture at the time of mixing (3 min) is extrapolated backward from the data. We do this graphically. The cold water is initially in the calorimeter and therefore the calorimeter undergoes the same temperature change as the cold water. A graph showing the idealized behavior of the system is shown below. The temperature axis is intentionally unlabeled and the data relevant to the determination of the calorimeter constant are in the table below the graph. In the following questions, as in the laboratory, the heats and enthalpies in the exercise vary over a wide range. Almost all data are limited to 3 significant figures. To insure that your answers are properly interpreted, please give your answers in the units indicated in the questions but do not include the units with the answer you submit. What would the final temperature at the time of mixing, T_mix, have been if the calorimeter had been ideal (C_cal = 0)? Enter Your Answer: How much heat (in kJ) was lost by the hot water? (Pay attention to the sign!) Enter your Answer: How much heat (in kJ) was gained by the cold water? Enter your Answer: What is the calorimeter constant (in J/degree C)? Enter Your Answer: The calorimeter above was used to determine the enthalpy of the reaction between hydrochloric acid and sodium hydroxide. The relevant temperatures at the time of reaction (assuming the reaction was instantaneous) were determined as above. Temperatures were recorded as a function of time and extrapolated to the time of mixing. The generic graph is shown below. The temperature axis is again intentionally not displayed. The relevant information is given in the table below.

Explanation / Answer

Calorimeter

1. Let Tf be the final temperature

then,

47.7 (Tf - 19.1) = 54.1 (47.4 - Tf)

47.7Tf - 911.07 = 2564.34 - 54.1Tf

Tf = 3475.41/101.8 = 34.14 oC

2. Heat lost by hot water = 54.1 x 4.186 (47.4 - 32.8) = 3.306 kJ

3. Heat gained by cold water = 47.7 x 4.186 (32.8 - 19.1) = 2.735 kJ

4. Heat gained by calorimeter = 3.306 - 2.735 = 0.571 kJ

Calorimeter constant = 0.571 x 1000/(32.8 - 19.1) = 41.68 J/oC

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