Heat of Neutralization 28 Pre-lab Questions Before beginning this experiment in
ID: 1004407 • Letter: H
Question
Heat of Neutralization 28 Pre-lab Questions Before beginning this experiment in the laboratory, you should be able to answer the following questions. Define endothermic and exothermic reactions in terms of the sign of AH. 2. A720 mL sample of water was cooled from 50.0 C to 10.0 C. How much heat was lost? Define the term heat capacity. 4 How many joules are required to change the temperature ofso.o gof water from 21.3 oc to 43.5 ec? 5. Define the term specific heat 6. Calculate the final temperature when so ml of water at 65 c are added to 25 ml of water at 25 C 7 Describe how you could determine the specific heat of a metal by using the apparatus and techniques from this experimentExplanation / Answer
1. If a reaction is endothermic that is the system that releases heat to the surroundings, an exothermic reaction, has a negative H by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system.
In Endothermic reactions A system of reactants that absorbs heat from the surroundings in an endothermic reaction has a positive H, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.
2.
Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g oC)
T = change in temperature = Tfinal - Tinitial in oC
Q = 720 x 4.19 x (50-10)
Q = 120672 Jouels or 120.672 Kilo Joules of heat was lost.
3. The heat capacity of a defined system is the amount of heat (usually expressed in calories, kilocalories, or joules) needed to raise the system's temperature by one degree (usually expressed in Celsius or Kelvin).
4. Use the same formuls used in problem 2
Q = 50 x 4.19 x (43.5-21.3)
Q = 4650.9 Joules of heat is required
5. The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.
6.
Heat lost by 50 ml water = heat gained by 25 ml water .
50 X 4.19 x (65 -Tf) = 25 x 4.19 x (Tf - 25)
13617.5 - 209.5Tf = 104.75 Tf - 2618.75
16236.25 = 314.25 Tf
Tf = 51.66 oC
Final Temperature is 51.66 oC.
7.
The amount of heat energy involved in changing the temperature of a sample of a particular substance depends on three parameters:
T = final temperature initia ltemperature
The amount of heat energy that is transferred, q, in the process of producing a temperature change can be calculated according to the following equation:
q = s X m X T
In this experiment, you will determine the specific heat of a metal. A heated sample of this metal will be poured into a calorimeter, consisting of cool water contained in a plastic foam cup. Shortly after mixing, the water and the metal will have come to the same temperature. Because plastic foam is a good insulator, heat cannot easily escape from the calorimeter to the surroundings. Therefore, the heat lost by the metal can be said to be equal to the heat gained by the water.
The amount of heat energy gained by the water will be calculated in the following manner:
The specific heat of water is known. The temperature changes of the water, and of the metal, can be measured, as can the mass of the water and the mass of the metal. Using this data, the specific heat of the metal can be calculated using equation #3 above.
T (metal) = ________ Co
Heat energy gained by the water. = ________ calories
s (metal) = specific heat (water) X mass (water) X T (water) / mass (metal) X T (metal)
Specific heat of the metal = ________ cal / g • Co
The amount of heat energy that is required to raise the temperature of one gram of a substance by one degree Celsius is called the specific heat capacity, or simply the specific heat, of that substance. Water, for instance, has a specific heat of 1.0 cal / g • Co.The amount of heat energy involved in changing the temperature of a sample of a particular substance depends on three parameters:
- the specific heat of the substance = s
- the mass of the sample = m
- the magnitude of the temperature change = T
T = final temperature initia ltemperature
The amount of heat energy that is transferred, q, in the process of producing a temperature change can be calculated according to the following equation:
q = s X m X T
In this experiment, you will determine the specific heat of a metal. A heated sample of this metal will be poured into a calorimeter, consisting of cool water contained in a plastic foam cup. Shortly after mixing, the water and the metal will have come to the same temperature. Because plastic foam is a good insulator, heat cannot easily escape from the calorimeter to the surroundings. Therefore, the heat lost by the metal can be said to be equal to the heat gained by the water.
The amount of heat energy gained by the water will be calculated in the following manner:
- heat gained (water) = specific heat (water) X mass (water) X T (water)
- heat lost (metal) = specific heat (metal) X mass (metal) X T (metal)
- Because the heat gained must equal the heat lost: specific heat (water) X mass (water) X T (water) = specific heat (metal) X mass (metal) X T (metal)
The specific heat of water is known. The temperature changes of the water, and of the metal, can be measured, as can the mass of the water and the mass of the metal. Using this data, the specific heat of the metal can be calculated using equation #3 above.
Materials: balance, weighing dish, 250 ml beaker, 400 ml beaker, 100 ml graduated cylinder, 25 x 150 mm test tube, stirring rod, utility clamp, ring stand, ring support, wire gauze, gas burner, plastic foam cup, 2 thermometers, metal shot, paper towel, distilled water Experimental Design:- Fill a 250 ml beaker about 3/4 full of water. Place the beaker on a wire gauze on a ring support clamped to a ring stand. Use a gas burner to bring the water to a slow boil. While the water is heating, proceed to step 2.
- Using your weighing dish, obtain a sample of the metal being used that will fill the test tube about 1/4 full. Find the mass of the metal shot to the nearest 0.01 g and record on the data table.
- CAREFULLY transfer the metal shot to a large, dry test tube. Be careful to pour the metal shot into the tube slowly so that the bottom of the test tube is not broken in this process. Suspend the test tube in the boiling water with a utility clamp. Position the test tube so that the metal shot is below the level of water in the beaker (be sure the bottom of the test tube does not touch the bottom of the beaker). Adjust the flame so the water is just boiling gently. Allow the test tube to remain in the boiling water bath for at least 10 minutes. Proceed to step 4 while the metal shot is heating.
- Carefully measure out 100.0 ml of distilled water in a graduated cylinder, and pour the water into a plastic foam cup. Place the cup in a 400 ml beaker for support. Place a thermometer and stirring rod in the cup. Record on the data table the mass of the 100.0 ml water sample, REMEMBER that 1 milliliter of water has a mass of 1 gram.
- After the metal shot has been heating for at least 10 minutes, using the other thermometer, measure the temperature of the hot water bath. The temperature of the metal shot is the same as the water bath. Record this temperature to the nearest 0.5 oC as the initial temperature of the metal sample in the data table. Read the temperature of the water in the calorimeter to the nearest 0.5 oC and record in the data table as the initial temperature of the water.
- Remove the test tube from the bath, using the clamp as a holder. Carefully, but quickly, pour the metal shot into the water in the plastic cup (use a paper towel to keep any hot water on the tube from dropping into the calorimeter). Use the stirring rod to gently stir the metal shot (do not stir the shot with the thermometer). Note the temperature frequently. As the temperature begins to change more slowly, watch the thermometer continuously so as not to miss the maximum temperature reached. Record this maximum temperature on the data table to the nearest 0.5 oC, as the final temperature of water and metal.
- Calculate the T (water) and T (metal). These are just simple subtractions. Remember, the water and metal had the same final temperature, but different initial temperatures. Also remember that T is never negative, so one calculation will be temperature (final) temperature (initial), and the other will be temperature (initial) temperature (final).T (water) = ________ Co
T (metal) = ________ Co
- Calculate the heat energy gained by the water. Use formula (1) from the introduction for this calculation.
Heat energy gained by the water. = ________ calories
- Remember that the heat gained by the water is equal to the heat lost by the metal, calculate the specific heat of the metal. This is simply rearranging formula (3) from the introduction to solve for the specific heat of the metal, and plugging-in your data. The specific heat of water is again, 1cal / g • Co The formula would be:
s (metal) = specific heat (water) X mass (water) X T (water) / mass (metal) X T (metal)
Specific heat of the metal = ________ cal / g • Co
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