Experiment 3 - Lab Report PDF

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Objective: The objective of this experiment is to study the heats involved in the change of water from a solid to a liquid and from a liquid to a gas by the method of mixtures.

Apparatus: 1. 2. 3. 4. 5. 6. 7. 8. 9.

Calorimeter Boiler Water trap Rubber tubing Triple-beam balance 0-100ºC thermometer 0-50ºC thermometer Ice Paper towels

Theory: The amount of heat required to change a unit mass of a substance from the solid to the liquid state without a change in temperature is called the latent heat of fusion of the substance. The latent heat of vaporization is the amount of heat required to change a unit mass of a substance from the liquid to the vapor state without a change in temperature. The experimental determination of the heat of fusion of ice and heat of vaporization of water is made by the method of mixtures.

Procedures: 1. Weigh the empty calorimeter. 2. Fill the calorimeter half full of water about 10 above room temperature and weigh it. Replace it in the outer calorimeter jacket and record the temperature of the water. 3. Dry some small pieces of ice on paper towel and add them to the water without touching the ice with your fingers. Add the ice until the temperature is about 10 below room temperature, keeping the mixture well stirred with the 0-50ºC thermometer. Record the equilibrium temperature when the ice is entirely melted. Weigh the calorimeter with its contents again, but without thermometer. 4. Fill the calorimeter three quarters full of water about 15 below room temperature and again weigh it. Replace it in the outer calorimeter jacket. 5. Set up steam generator. Record the temperature of the cold water and quickly immerse the steam tube. Allow the steam to pass in and condense until the temperature of the water is about 15 above room temperature, stirring continuously with the 0-100C thermometer. Remove the steam tube and record the equilibrium temperature. Weigh the calorimeter with its contents again, but without the thermometer.

Data:

Mass of calorimeter: 33.9g Specific heat of calorimeter: 0.215 Heat of Fusion: Mass of calorimeter and water Mass of water Initial temperature of water Equilibrium temperature Mass of calorimeter, water, and melted ice Mass of ice Calculated heat of fusion of ice Accepted value of heat of fusion of ice Percent error

Heat of Vaporization: Mass of calorimeter and water Mass of water Initial temperature of water Equilibrium temperature Mass of calorimeter, water, and condensed steam Mass of steam Calculated heat of vaporization of water Accepted value of heat of vaporization of water Percent error

183.7 g 149.9 g 35ºC 14ºC 222.6 g 38.9 g 70.85 cal/g 79.7 cal/g 11%

259.3 g 225.5 g 10ºC 39ºC 271.1 g 11.8 g 511 cal/g 540 cal/g 5.4%

Calculations: 1. From the data of Procedures 1-3, compute the heat of fusion of ice. (mcw +m 1c 1)(t 1−t 2)– Mcw (t 2−0) M ( 149.9∗1+33.9∗0.215 )( 35 −14 )−38.9(14−0) = 38.9 = 70.85 cal/g

Lf =

2. From the data of Procedures 4 and 5, compute the heat of vaporization of water.

( mcw+m 1 c 1 ) ( t 2− t 1 )−Mcw (100−t 2) M ( 225.5 + 33.9∗0.215 )(39− 10 )−11.8 (100−39 ) = 11.8 = 511.1 cal/g

Lv =

3. Compare your results for the heat of fusion and the heat of vaporization with the known values by finding the percent error. Percent error for heat of fusion = (70.85 – 79.7) / (79.7) *100 = 11% Percent error for heat of vaporization = (511-540) / (540) *100 = 5.4%

Post Lab Questions: 1. Discuss the principal sources of error in this experiment. The principal sources of error in this experiment could be inaccurate reading of the thermometer, heat loss, calorimeter not being well insulated, and not stirring the mixture continually. 2. Why is it necessary to stir the mixtures? It is necessary to stir the mixtures as it gives a better accurate reading of the calorimeter and allows the water to reach equilibrium temperature uniformly. 3. What error would have been introduced if the ice had not been dry? If the ice had not been dry then it would have caused an increase in final temperature and consequently the heat of fusion would have been lower. 4. Suppose the steam generator produced steam so fast that some of it bubbled up through the water and escaped from the calorimeter without condensing. Would this cause an error in your results? Explain. If the steam bubbled up and escape from the calorimeter without condensing then it would have caused an error in the results as we would not been up able to measure the mass of the steam and consequently it would have increase the heat of vaporization. 5. In the Theory section of this experiment it was pointed out that Equation 20.1 assumed the ice was at 0C before it was put in the water and that, similarly, Equation 20.2 assumed the entering steam to be at 100C. (a) How good do you think these assumptions are? (b) What should you have done to make sure they’re very good? (c) How should Equations 20.1 and 20.2 be modified if the above assumptions are not good- that is, if the ice is much colder than 0C and the steam much hotter than 100C when they are put in the water? (d) Could you run the experiment on this basis? a) The assumptions are good because we know that the boiling temperature is 100ºC and the freezing temperature of water is 0ºC. b) The assumptions are great so no improvements needed. c) If the ice is much colder than 0ºC then equation 20.1 would be modified by having 0 minus the new temperature of the ice. If the steam is much hotter than 100C then equation 20.2 is changed by having the new steam temperature minus 100. d) Yes, the experiment can be run on this basis but it would be a little more difficult. 6. (a) What is the purpose of the water trap in determining the heat of vaporization? The purpose of the water trap is to prevent the condensation from going into the water steam. (b) How would its absence have affected the results?

If the water trap was not there then the mass of the steam would be greater than the actual mass. 7. The Theory section of this experiment suggests that when a liquid starts to boil, the formation of bubbles should begin near the surface. Why do the bubbles actually form first at the bottom of the containing vessel? The bubbles form first at the bottom of the containing vessel because these bubbles are the air that was dissolved in the water and they stay at the bottom because the vapor pressure is different from the atmospheric pressure at the time when the temperature of the water is not hot enough. 8. A glass tumbler of 200g mass and 5 cm inside diameter is filled to a depth of 10 cm with tap water whose temperature is 10 C and 50 g of ice at 0 C are then added. At the end of an hour, it is noted that the ice has just finished melting. What was the average rate, in calories per second, at which heat flowed into the glass from its surroundings? Mass of water=πr2hρ=196.35g Total heat provided by water to ice=mCpΔT=8219J Total heat needed to melt ice=mL=16700J Extra heat required=8481J=2026Cal Heat flow rate=2026/3600Cal/s=0.5628Cal/s

Summary and Discussion: This experiment demonstrated the changing of a substance from the solid to the liquid state and from the liquid to the gaseous state. It gave an understanding of how to find the heat of fusion and vaporization. Comparing the results, we got from the experiment to the actual values we got 11% and 5.4% as our percent errors. So not bad for our errors, the errors we got could have been because the calorimeter did not do good job at insulating and there could have been heat loss due to not enough stirring the mixture....