SPECIFIC HEAT OF WATER LAB 11-2

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Marcia Berry
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1 CONCEPT Heat of Fusion Changes of state (phase changes) involve the conversion or transition of matter from one of the common states (solid, liquid or gas) to another. Examples include fusion or melting (change from the solid to the liquid state), freezing (change from the liquid to the solid state), evaporation or vaporization (change from the liquid to the gaseous state), condensation (change from the gaseous state to the liquid state), sublimation (change from the solid directly to the gaseous state), and deposition (change from the gaseous state directly to the solid state). Heat of Fusion is the amount of heat released as 1 kilogram of liquid freezes (solidifies or crystallizes), or the amount of heat absorbed as a solid melts (fuses or liquefies). The exact value of the Heat of Fusion is different for different substances. For water, the Heat of Fusion is 335 kilojoules per kilogram, or 335 Joules per gram, also expressed as 80 calories per gram. Although the Joule is the official SI unit for measurement of energy, it will be simpler to calculate in this experiment if we use the calorie as an energy unit. 1 calorie = Joules. The calorie is simple to use in problems involving water, because 1 calorie of heat energy will raise the temperature of 1 gram of water by 1 o C. Typically, changes of state of a pure substance, such as melting, freezing, evaporation, condensation, etc., occur at a constant temperature. Temperature change only occurs when all of the phase changes are completed. This means that during the phase change, energy is being absorbed or emitted with no change in temperature of the material being studied. This energy can be absorbed or supplied by some other substance in the environment, and absorbing or supplying that energy can alter the temperature of that material. Specific Heat When a substance changes temperature, but does not change its state, its temperature goes up by an amount proportional to the amount of heat energy absorbed or lost. The specific amount of energy that must be lost or gained to change the temperature of a standard mass (1 kilogram for physics) of a substance by 1 o Celsius is called the Specific Heat. The exact value of the specific heat varies, depending on the particular substance. For water, the Specific Heat is 4.2 kilojoules per kilogram per o Celsius, or 4.2 joules per gram per o Celsius, also expressed as 1.0 calories per gram per o C.

2 Application of Heat of Fusion and Specific Heat in this experiment In our study today, we are going to study the Heat of Fusion of water. We are going to place some solid water (ice) at its melting temperature of ~0 o C into some warm liquid water that has been heated to a warmer temperature than the lab room. By doing this procedure in an insulated container called a calorimeter, we will limit transfer of heat to between the ice and the water. Because of the insulation, very little heat will be transferred between the water or ice and the environment. The calorimeter simply consists of a small styrofoam cup within a larger styrofoam cup. An insulating cardboard or styrofoam cover is pierced by a hole to allow insertion of a thermometer, which also serves as a stirring device. The insulation (styrofoam, cardboard and air spaces) of the calorimeter will prevent heat transfer to any other substance or material in the environment. By assuming that the heat absorbed by the styrofoam calorimeter and air is approximately zero, we can approximate the heat absorbed by the ice as it melts as being equal to the heat lost by the water as it cools off. Heat gained = Heat lost equation 1 The heat gained by the ice as it melts is calculated by multiplying the mass of the ice times the heat of fusion of ice. Heat gained m ( H ) equation 2 ice f Once the ice melts, the cold water from the melted ice (which will be at the same temperature as the ice) will gain some heat from the warm water in the calorimeter. The heat gained by the cold water is calculated by multiplying the mass of the ice times the specific heat of water times the rise in temperature of the cold water. (See Text, p. 97.) Heat gained ( ice water) m ( H )( T T ) equation 3 ice sp ice final The total heat gained is equal to the sum of equations 2 and 3. Total Heat gained m ( H ) H ( T T ) equation 4 ice f sp ice final The heat lost by the warm water in the calorimeter will be calculated by multiplying the mass of the warm water times the specific heat of water times the drop in temperature. Heat lost ( warm water) m ( H )( T T ) equation 5 water sp w final The total heat lost is equal to equation 4. Since heat lost equals heat gained (equation 1), then equation 5 equals equation 4. m ( H ) ( m )( H )( T T ) ( m )( H )( T T ) equation 6 ice f ice sp ice final water sp water final

Equation 6 can be solved for the Heat of Fusion. H f ( mwater )( H sp )( Twater Tfinal ) ( mice)( Tice Tfinal ) equation 7 m ice OBJECTIVE Investigate the Heat of Fusion of water.

3 Equation 6 can be solved for the Heat of Fusion. H f ( mwater )( H sp )( Twater Tfinal ) ( mice)( Tice Tfinal ) equation 7 m ice OBJECTIVE Investigate the Heat of Fusion of water. MATERIALS 2 Styrofoam cups distilled water Beakers Beaker tongs Hot plate 2 thermometers String Stopwatch Electronic balance PROCEDURE Specific heat of water 1. Place some solid water (ice) at its melting temperature of ~0 o C into some warm liquid water that has been heated to a warmer temperature than the lab room. By doing this procedure in an insulated container called a calorimeter, we will limit transfer of heat to between the ice and the water. Because of the insulation, very little heat will be transferred between the water or ice and the environment. The calorimeter simply consists of a small styrofoam cup within a larger styrofoam cup. An insulating cardboard or Styrofoam cover is pierced by a hole to allow insertion of a thermometer, which also serves as a stirring device. The insulation (styrofoam, cardboard and air spaces) of The calorimeter will prevent heat transfer to any other substance or material in the environment. By assuming that the heat absorbed by the styrofoam calorimeter and air is approximately zero, we can approximate the heat absorbed by the ice as it melts as being equal to the heat lost by the water as it cools off. 2. Determine the room temperature by setting the thermometer on the table for 5 minutes, and then reading it. Record the temperature in o F, as it is displayed, then calculate and record the temperature in o C to the same number of decimal places as the thermometer reading.

4 3. Fill the 400 ml beaker about half full of distilled water from the special supply bottle. 4. Place the beaker of water on the hot plate and heat to a temperature about 15 o C (27 o F) above room temperature. If the water becomes too hot, you can cool it down by adding room temperature distilled water from the supply bottle. 5. Assemble the inner and outer calorimeter cups and put on the cardboard cover. 6. Weigh the assembled calorimeter, and record the weight. 7. Fill the inner calorimeter cup about half full of warm water from the beaker on the hot plate. 8. Re-cover the calorimeter and weigh the combined cup and the water, and record the mass. 9. Insert the thermometer into the ice container. 10. Allow the temperature to stabilize, a few minutes, and then read the thermometer, recording the temperature as Tice. 11. Insert the thermometer into the calorimeter, and stir briefly. 12. When the temperature stabilizes, read the thermometer, estimating between the lines to the nearest 0.1 o C, recording the temperature of the warm water as Tw. 13. Acquire a tablespoon or so of ice, dry the pieces of ice with paper towels, and dump the dried ice into the calorimeter. Be careful not to touch the ice with your bare fingers or hands, as this will cause melting and create errors in your results. 14. Quickly cover the calorimeter and stir with the thermometer, until all the ice has melted. If necessary, add more dried ice until the temperature reaches at least 10 o C below room temperature. 15. When all the ice has melted, and the temperature reaches a low point, around 5-10 o C (40-50 o F), record the temperature, as Tfinal. 16. Remove the thermometer and weigh the covered calorimeter and cool water. Record the mass. 17. Subtract the mass of the calorimeter from the mass of the calorimeter and warm water. Record as the mass of the warm water, mw. 18. Subtract the mass of the calorimeter and warm water from the mass of the calorimeter and cold water. Record as the mass of the ice, mice. 19. Use equation 7 to calculate the Heat of Fusion of water. 20. Repeat steps 6, 7, and 11 through 18 until you obtain 3 non-deviant values for Hf. 21. Add the three values of Hf to obtain a total. 22. Divide the total obtained in step 20 by three to obtain the average. 23. Subtract the average experimental value of Hf for water, obtained in step 21, from the accepted standard value of 80.0 calories per gram, divide by the standard and multiply times 100 to obtain the per cent difference between your value and the standard. standard - experimental 80.0 average % difference x100% x100% s tan dard 80.0 Be sure and mention this per cent difference in the Results and Discussion sections of your lab report. 24. In the Discussion section of your lab report, state your conclusion by answering the following question: Did your experimental determination of the Heat of Fusion of water agree with, or come close to the accepted standard value? Explain why or why not.

5 DATA PART I Vol, of tap (ml) Mass of tap Temp. of tap Vol. of warm (ml) Mass of warm Temp. of warm Final temp. of Mixture PART II Aluminum Brass Copper Iron (steel) Lead Mass of Temp. of Mass of Initial Temp. of Final Temp. Specific Heat Joules/kg.o C % Diff. CALCULATIONS PART I Heat lost by hot water = heat gained by tap water m c ( T T ) m c ( T T ) hot hot hot final tap tap final tap PART II Heat lost by metal sample = heat gained by water (assume no heat gained by cup) m c T m c T metal metal metal water water water DATA ANALYSIS 1. Does the final temperature reached in part I seem reasonable? Explain, using the word heat in your explanation. 2. If you mixed the hot water with alcohol would you obtain the same results? Explain your answer. 3. Explain qualitatively what is happening to account for the final temperature. 4. In which direction, i.e., from which object did heat flow when i. The metal was placed in boiling water? ii. The metal was placed in the cool water? 5. Calculate the % error of your results with the expected. Explain the source of your error. 6. It is not likely that the temperature of your boiling water was exactly 100 o C. Can you suggest several factors that would affect the boiling point of water? 7. What is thermal equilibrium? How can you recognize when thermal equilibrium has been reached? 8. Why was the metal left in the boiling water for 10 minutes? 9. Why the cool water was purposely started well below room temperature? 10. Define specific heat. Do not copy the definition from the textbook. 11. Explain why water is used in heating systems for buildings

SPECIFIC HEAT CAPACITY AND HEAT OF FUSION

SPECIFIC HEAT CAPACITY AND HEAT OF FUSION Apparatus on each table: Thermometer, metal cube, complete calorimeter, outer calorimeter can (aluminum only), balance, 4 styrofoam cups, graduated container,