# q = (mass) x (specific heat) x T = m c T (1)

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1 Experiment: Heat Effects and Calorimetry Heat is a form of energy, sometimes called thermal energy, which can pass spontaneously from an object at a high temperature to an object at a lower temperature. If the two objects are in contact they will, given sufficient time, both reach the same temperature; we call this thermal equilibrium. Heat flow is ordinarily measured in a device called a calorimeter. A calorimeter is simply a container with insulating walls, made so that essentially no heat is exchanged between the contents of the calorimeter and the surroundings. Within the calorimeter chemical reactions may occur or heat may pass from one part of the contents to another, but no heat flows into or out of the calorimeter from or to the surrounding. A. Specific Heat. When heat flows into a substance the temperature of that substance will increase. The quantity of heat, q, required to cause a temperature change, T, of any substance is proportional to the mass, m, of the substance and the temperature change, as shown in equation (1). The proportionality constant is called the specific heat of that substance. q = (mass) x (specific heat) x T = m c T (1) The specific heat can be considered to be the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (if you make m and T in equation 1 both equal to one, then q will equal specific heat, c). Amounts of heat are measured in either joules or calories. TO raise the temperature of one gram of water by one degree Celsius 4.18 joules of heat must be added to the water. The specific heat of water is therefore 4.18 J/g C. Since 4.18 joules equals one calorie, we can say that the specific heat of water is 1.00 calories/g C. Ordinarily heat flow into or out of a substance is determined by the effect which that flow has on a known amount of water. Because water plays such an important role in these measurements the calorie, which was the unit of heat most commonly used until recently, was actually defined to be equal to the specific heat of water. The specific heat of a metal can readily be measured in a calorimeter. A weighed amount of metal is heated to some known temperature and is then quickly poured into a calorimeter that contains a measured amount of water at a known temperature. Heat flows from the metal to the water, and the two equilibrate at some temperature between the initial temperatures of the metal and the water. Assuming that no heat is lost from the calorimeter to the surroundings, and that a negligible amount of heat is absorbed by the calorimeter walls, the amount of heat that flows from the metal as it cools is equal to the amount of heat absorbed by the water. In thermodynamic terms, the heat flow for the metal is equal in magnitude but opposite in direction, and hence in sign, to that for the water. For the heat flow, q, q metal = -q water (2) If we now express heat flow in terms of equation 1 for both the water and the metal M, we get C M m M T M = C H2O m H2O T H2O (3) In this experiment we measure the masses of water and metal and their initial and final temperatures. (Note that T M <0 and T H2O.), since T = T final T initial.) Given the specific heat of water we can find the positive specific heat of the metal by equation 3. We will use this procedure to obtain the specific heat of an unknown metal.

2 The specific heat of a metal is related in a simply way to its atomic mass. Dulong and Petit discovered many years ago that about 25 joules were required to raise the temperature of one mole of many metals by one degree Celsius. This relation, shown in equation 4, is known as the Law of Dulong and Petit: AM = 25 (4) C metal (J/g C) Where AM is the atomic mass of the metal. Once the specific heat of the metal is known, the approximate atomic mass can be calculated by equation 4. The Law of Dulong and Petit was one of the few rules available to early chemists in their studies of atomic masses. (Note: 25 is an approximate, not exact #.) B. Heat of Solution. When a chemical reaction occurs in water solution, the situation is similar to that which is present when a hot metal sample is put into water. With such a reaction there is an exchange of heat between the reaction mixture and the solvent, water. As in the specific heat experiment, the heat flow for the reaction mixture is equal in magnitude but opposite in sign to that for the water. The heat flow associated with the reaction mixture is also equal to the enthalpy change, H, for the reaction, so we obtain the equation q reaction = H reaction = -q H2O (5) By measuring the mass of the water used as a solvent, and by observing the temperature change that the water undergoes, we can find q H2O by equation 1 and H by equation 5. If the temperature of the water goes up, heat has been given off by the reaction mixture, so the reaction is exothermic; q H2O, is positive and H is negative. If the temperature of the water goes down, the reaction mixture has absorbed heat from the water and the reaction is endothermic. In this case, q H2O is negative and H is positive. Both exo- and endothermic reactions are observed. One of the simplest reactions that can be studied in solution occurs when a solid is dissolved in water. As an example of such a reaction note the solution of NaOH in water. NaOH (s) Na + (aq) + OH - (aq); H = H solution (6) When this reaction occurs, the temperature of the solution becomes much higher than that of the NaOH and water that were sued. If we dissolve a known amount of NaOH in a measured amount of water in a calorimeter, and measure the temperature change that occurs, we can use equation 1 to find q H2O for the reaction and use equation 5 to obtain H. Noting that H is directly proportional to the amount of NaOH used, we can easily calculate H solution for either a gram or a mole of NaOH. In the second part of this experiment you will measure H solution for an unknown ionic solid. EXPERIMENTAL PROCEDURE Wear your safety glasses while performing this experiment: Part A Specific Heat of a Metal. From your instructor obtain a calorimeter, a sensitive thermometer, a sample of metal in a large stoppered test tube. The calorimeter consists of two nested expanded polystyrene coffee cups fitted with a Styrofoam cover. There are two holes in the cover for a thermometer and a glass stirring rode with a loop bent on one end. Assemble the experimental setup as shown in figure 1. Fill a 400-mL beaker two-thirds full of water and begin heating it to boiling. Fill a 100-mL beaker two-thirds full of water and begin heating it to boiling a well. While the water is heating, weight your sample of unknown metal in the large stoppered test tube to the nearest 0.1g on a top loading balance. Pour the metal into a dry container and weigh the empty test tube and stopper. Replace the metal in the test tube and put the LOOSELY stoppered tube into the hot water in the400-ml beaker. The water level in the 400-mL beaker should be high enough so that the top of the metal is below the water surface. Continue heating the metal in the water for at least 5 minutes after the water begins to boil to ensure that the metal attains the temperature of the boiling water. Add water as necessary from your 100-mL beaker to the 400-mL beaker to maintain the water level.

4 Experiment: Calorimetry Name: Block: Pre-lab Heat Effects and Calorimetry (To be handed in before starting the experiment.) SHOW SET-UPS FOR CREDIT. 1. A metal sample with a mass of 63.2 g. and at a temperature of C was placed in 41.0 g. of water in a calorimeter at 24.5 C. At equilibrium the temperature of the water and metal was 35.0 C. A. What was T for the water? ( T = T final T initial ) B. What was T for the metal? C. Taking the specific heat of water to be 4.18J/g C, calculate the specific heat of the metal, using equation 3. D. What is the approximate atomic mass of the metal? (Use equation 4.) 2. When 5.0 g of KNO 3 were dissolved in 49.0 g. water in a calorimeter at 24.0 C, the temperature of the solution fell to 15.6 C. A. Is this solution reaction exothermic? Why? B. What was T for the water? ( T = T final T initial ) C. Calculate q H2O, using equation 1. D. Find q for the reaction as it occurred in the calorimeter (equation 5). E. Find H solution in joules/g H = F. Find H solution in joules/mole H = G. Write the chemical equation that represents the heat of solution for KNO 3. H. Using enthalpies of formation, H f, given in the back of your text book (A19-A22) to calculate H solution for the reaction in part G. (Reminder: H rxn = Σ H f products - Σ H f reactants) I. Calculate your percent error using your answers to part F and H.

5 Experiment: Calorimetry Name: Block: Post-lab Heat Effects and Calorimetry Due: Day after lab. Part A Specific Heat of a Metal. 1. For the following errors describe how they would affect your calculated specific heat increase, decrease, remain the same). a. The metal did not reach thermal equilibrium with the boiling water. b. When pouring the metal into the distilled water, some hot water from the wet test tube fell into the distilled water as well.. c. Heat from the warming water was lost to the surroundings. d. The mass of the empty test tube was taken when the test tube was slightly wet. e. The initial temperature of the water was recorded higher than it actually was by reading the thermometer from above. Part B Heat of Solution. 1. We found the heat of solution of an ionic solid for part B of this lab. Describe heat of solution briefly in your own words. 2. If we were to graph the entire class s results for this lab, would we likely have a. Mostly low results b. Mostly high results c. Approximately equal number of low and high results 3. Given your answer to #2, did we have random or systematic error? 4. For the following errors describe how they would affect your calculated heat of solution. a. Not all of the ionic solid was dissolved. b. The ionic solid dissolved slowly enough that much of the heat dissipated to the surroundings. c. The student recording the initial temperature of the distilled water did not wait for the thermometer to come to thermal equilibrium with the distilled water prior to recording the initial temperature. (The temperature of the room was warmer than was the distilled water.) d. The ionic solid had absorbed come water from the air prior to being dissolved in the distilled water. e. The balance read zero prior to placing the weigh boat on it. The student did not tare the mass of the weigh boat prior to adding the ionic solid. The balance read 6.00 grams with the weigh boat and solid on it. The student recorded the mass of the ionic solid as 6.00 grams.

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