Activity 201 5 Calorimetry Directions: This GLA worksheet goes over the concepts of heat and calorimetry. Part A introduces the concepts of heat and specific heat capacity. Part B introduces calorimetry and goes over simple calorimetry problems. Part C discusses the use of calorimetry to determine the enthalpy of a reaction.. See www.canyons.edu/departments/chem/gla for additional materials. Part A Heat and Specific Heat Capacity Heat (q) is defined as the transfer of energy from one material to another due to differences in their thermal energy. This transfer of energy is accompanied by a change in the substance s temperature, which is defined the measurement of the thermal energy an object contains. For example, when a hot cup of coffee is sitting on a table it will transfer its thermal energy to the surrounding environment (coffee mug, table, hands holding the coffee, etc.). The surroundings will begin to feel warmer and the coffee will begin to feel colder. If the object (called the system) loses heat to anything it is in contact with (called the surroundings), the process is said to be exothermic and the sign of q is negative. When this occurs, the energy of the system decreases and the energy of the surroundings increases. If heat is transferred from the surroundings to the system, the process is said to be endothermic and the sign of q is positive. When this occurs, the temperature of the system increases and the temperature of the surroundings decreases. (Note that heat will never be a negative value. The sign of q only denotes direction of heat transfer.) Eventually, both will reach the same temperature and once this occurs, there is no net energy transfer and the two objects are said to have reached thermal equilibrium. The magnitude of the change in temperature of an object (system or surroundings) depends on its specific heat capacity. Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 C. It can be thought of as the amount of energy an object can hold before it begins to increase in temperature. It is an intrinsic property, meaning different materials will have their own unique and constant value for specific heat capacity. A good illustration of this is when we visit the beach on a hot summer day. The dry sand will feel much, much hotter than the ocean water. Why is that if they are both sitting under the same sun? It has to do with their specific heat capacity. Water has a high specific heat capacity (4.184 J/g C), so it will be able to hold a lot heat before it will begin to increase in temperature. Sand, on the other hand, has a relatively low specific heat capacity (0.840 J/g C), so it cannot hold as much heat, and will increase in temperature much more rapidly. Specific heat capacity and change in temperature are related by the following equation: q = mc s T where q represents heat, m represents mass, c s represents specific heat capacity, and T represents the change in temperature. (Recall that T = T final T initial.) Determine the heat (q) associated with a decrease in the temperature of 375 g of water from 31.5 C to 24.6 C. The specific heat capacity of water is 4.18 J/g C. Activity 201 5 Page 1 of 6
Since we are given, mass (m), specific heat capacity (c s), and a temperature change ( T), we can use the q = mc s T equation to solve for q. q = (375 g)(4.18 J/g C)(24.6 C 31.5 C) = (375 g)(4.18 J/g C)(-6.9 C) = -1.1 10 4 J The grams (g) and degrees Celsius ( C) units cancel so that heat has units of Joules (J). This process is exothermic, so 1.1 x 10 4 J was transferred from the system (the water) to the surroundings (everything else). Example #1 Determine the specific heat capacity of 250. g of an unknown metal. When the metal absorbs 18.1 kj of heat, its temperature increases from 23.1 C to 103.5 C. Part B Calorimetry and Thermal Equilibrium A technique used to measure the amount of heat transferred to or from a substance in a chemical or physical process is called calorimetry. A calorimeter is a device used to measure the heat transferred between a system and its surroundings by measuring the change in temperature. It is an insulated container, so we can assume any heat transfer that occurs is only between the system and the calorimeter and its contents (e.g. no heat is transferred to the air, table, my hand, etc.). Calorimeters usually contain water to absorb or release the heat for the chemical process. Therefore, to find the heat transfer to or from the water (q H2O) we can use the q=mc T expression. Figure 1. Diagram representation of a calorimeter. The system is losing heat to the surroundings in (a), causing the calorimeter to increase in temperature. The system is absorbing heat from the surroundings in (b), causing the calorimeter to decrease in temperature. (Source: Chemistry, published by Openstax College, 2016) If the system undergoes an exothermic process (Figure 1a), the calorimeter will increase in temperature. If the system undergoes an endothermic process (Figure 1b), the calorimeter will decrease in temperature. Since any heat that is transferred is only between the system and the calorimeter, we can employ the Conservation of Energy to write: q system + q calorimeter = 0 Activity 201 5 Page 2 of 6
This equation can also be re-written as: (mc T) system = -(mc T) surroundings In simple calorimetry problems, we can measure the heat transfer between two objects at different temperatures. A 85.6 g piece of gold (Au) is submerged in a calorimeter containing 100.0 g of H 2O, which is initially at 25.0 C. The two reach thermal equilibrium at 43.2 C. Assuming that the heat transferred is only between the gold and the water, what was the initial temperature of the gold? The specific heat capacity of water is 4.18 J/g C and that of gold metal is 0.129 J/g C. We can write our equation as: (mc T) Au = -(mc T) H2 O Since we are looking for T initial of the silver, we can include the T equation: (mc[t final T initial]) Au = -(mc[t final T initial]) H2 O At thermal equilibrium, the temperature of both objects is equal. Therefore, the final temperature of both the gold and water is 43.2 C. All we are missing is the initial temperature of the gold: (85.6 g)(0.129 J/g C)(43.2 C T initial) = -(100.0 g)(4.18 J/g C)(43.2 C 25.0 C) When we solve for T initial of the gold, we get 732 C. Example #2 A 25.0 g piece of metal alloy that is initially at 95.0 C is placed in 50.0 g of H 2O that is initially at 24.5 C. The two reach thermal equilibrium at 28.0 C. Determine the specific heat capacity of the alloy metal. Part C Reaction Calorimetry When a chemical reaction occurs, often energy will be absorbed (endothermic) or released (exothermic) by the system. Calorimetry can also be used to determine the enthalpy of a chemical reaction ( H rxn). Recall that H rxn is the amount of heat that is either absorbed or released during a chemical process (refer to GLA 201-5 Hess s Law and Enthalpy of Formation for more information). In this case, the system Activity 201 5 Page 3 of 6
is defined as the chemical components that participate in the chemical reaction. The surroundings include everything else, which often means the water in which the reactants are dissolved. The change in temperature of the calorimeter is associated with the heat that is either released or absorbed in a chemical reaction: q reaction + q calorimeter = 0 There are various types of calorimeters that can be used for different types of reactions. Coffee cup calorimeters and solution calorimeters operate at constant pressure and are typically used to measure the heat of reactions that occur in solution. Bomb calorimeters operate at constant volume and are used to measure the heat of reactions that are highly exothermic, such as combustion reactions. The heat capacity of bomb calorimeters is often calibrated, since many operate at constant mass. You will often see the heat of the calorimeter calculated using the following equation: q cal = C cal T where C cal is the heat capacity of the calorimeter with units of kj/ C. However, since enthalpy is an extrinsic property, then the heat transferred during a reaction will depend on the initial amount of reactants. For example, if 2.0 L of gasoline undergo combustion, it will produce less heat than if 10.0 L of gasoline undergo combustion. In order to determine the enthalpy of the reaction ( H rxn), the calculated heat (q reaction) must be divided by the moles of limiting reactant. When 1.0 g of fructose, C 6H 12O 6(s), is burned in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 C. If the heat capacity of the calorimeter and its contents is 9.90 kj/ C, what is the enthalpy of the combustion of 1 mole fructose? The temperature of the calorimeter increased due to the heat transfer of the reaction to the calorimeter. We must first calculate the heat that the calorimeter absorbed: q cal = C cal T = (9.90 kj/ C)(1.58 C) = 15.6 kj Since q cal = -q rxn, then q rxn = -15.6 kj If we would like to determine enthalpy of the combustion of 1 mole of fructose, we need to divide the heat of the reaction by the moles of fructose that reacted: 11. 00 gg ffffffffffffffff 11 mmmmmmmm = 00. 0000000000 mmmmmm ffffffffffffffff 111111. 1111 gg HH rrrrrr = qq rrrrrr mmmmmmmmmm ffffffffffffffff = 1111. 66 kkkk 00. 0000000000 mmmmmm = 22. 8888 111133 kkkk/mmmmmm When 50.0 ml of 1.0 M HCl(aq) and 50.0 ml of 2.0 M NaOH(aq), both at 22.0 C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.7 C. What is the enthalpy of the reaction? The solution has a density of 1.03 g/ml and the specific heat capacity of water is 4.18 J/g C. Activity 201 5 Page 4 of 6
HCl(aq) + NaOH(aq) H 2O(l) + NaCl(aq) In this example, both reactants are dissolved in water, so any heat from the reaction is transferred to the surrounding water. The temperature change of the water is measured, so we can calculate the amount of heat that is absorbed by the water, which is equal (but opposite in sign) to the heat released by the reaction. 111111. 00 mmmm ssssssssssssssss 11. 0000 gg 11 mmmm = 111111 gg ssssssssssssssss q solution = mc solution T = (103 g)(4.18 J/g C)(28.7 C 22.0 C) = 2885 J Since q solution = -q rxn, then q rxn = -2885 J We now need to determine the enthalpy of the reaction, which means that we need to divide the heat by the moles, as in the previous example. However, in the previous example, fructose was burned in a combustion reaction, making O 2 the excess reactant and fructose the limiting reactant. In this example, we are presented with two reactants, so we will need to first determine the limiting reactant (refer to the Chem 151-9 Introduction to Stoichiometry and Chem 201-2 Stoichiometric Calculations with Excess Reactants GLAs for additional guidance): 00. 00000000 LL HHHHHH 11. 00 mmmmmmmm HHHHHH 11 LL HHHHHH 11 mmmmmmmm NNNNNNNN 11 mmmmmmmm HHHHHH = 00. 00000000 mmmmmmmm NNNNNNNN 00. 00000000 LL HHHHHH 22. 00 mmmmmmmm NNNNNNNN 11 LL NNNNNNNN 11 mmmmmmmm NNNNNNNN = 00. 111111 mmmmmmmm NNNNNNNN 11 mmmmmmmm NNNNNNNN The limiting reactant is HCl, meaning that we will need to divide by the moles of HCl to determine the enthalpy of the reaction: HH rrrrrr = qq rrrrrr mmmmmmmmmm ffffffffffffffff = 22222222 JJ 00. 00000000 mmmmmm = 55. 7777 111144 JJ/mmmmmm Example #3 When 3.12 g of glucose, C 6H 12O 6, is burned in a bomb calorimeter, the temperature of the calorimeter increases form 23.8 C to 35.6 C. The calorimeter contains 755 g of water, and the bomb itself has a heat capacity of 0.893 kj/c. What is the enthalpy of combustion of glucose? The heat capacity of water is 4.18 J/g C. Activity 201 5 Page 5 of 6
Part D Extra 1. A piece of an unknown metal weighs 550. g and requires 9.315 kj of energy to increase its temperature from 22.0 C to 65.2 C. What is the specific heat capacity of the metal? 2. What is the final temperature of a 275 kg sample of water, initially at 19.5 C, if 36.5 kj of heat were added to it? The specific heat capacity of water is 4.18 J/g C. 3. A 45.0 g piece of aluminum, which is initially at 19.5 C, is placed in 180.0 g of water, which is initially at 95.1 C. What is the final temperature of the mixture once the aluminum and water reach thermal equilibrium? The specific heat capacity of water is 4.18 J/g C and that of aluminum is 0.24 J/g C. 4. If a reaction produces 1.506 kj of heat, which is contained in 30.0 g of water that is initially at 26.5 C, what is the resulting temperature of the water? The specific heat capacity of the water is 4.18 J/g C. 5. When 1.025 g of naphthalene, C 10H 8, burns in a bomb calorimeter, the temperature rises from 24.25 C to 32.33 C. Find the enthalpy of combustion of naphthalene. The heat capacity of the calorimeter was determined to be 5.11 kj/ C. 6. Zinc metal reacts with hydrochloric acid according to the balanced equation: Zn(s) + 2 HCl(aq) ZnCl 2(aq) + H 2(g) When 0.110 g of Zn(s) is combined with 50.0 ml HCl solution in a coffee cup calorimeter, the temperature of the solution increased from 22.5 C to 23.8 C. Assuming an excess of HCl(aq), find the enthalpy of the reaction. The density of the solution is 1.03 g/ml and its specific heat capacity is 4.18 J/g C. 7. When 0.740 g of trinitrotoluene (TNT), C 7H 5N 2O 6, is burned in a bomb calorimeter, the temperature increases from 23.4 C to 26.9 C. The heat capacity of the calorimeter is 534 J/ C, and it contains 675 g of water. What is the enthalpy of combustion of TNT? 8. When 50.0 g of 0.200 M NaCl(aq) is added to 100.0 g of 0.200 M AgNO 3(aq), both initially at 24.1 C, in a calorimeter, the temperature increases to 25.2 C as AgCl(s) forms according to the chemical equation below. If the specific heat of the solution and products is 4.10 J/g C, calculate the enthalpy of the reaction ( H rxn). (The density of the solution is 1.00 g/ml.) NaCl(aq) + AgNO 3(aq) AgCl(s) + NaNO 3(aq) 9. The addition of 3.15 g of Ba(OH) 2 8H 2O to a solution of 1.52 g of NH 4SCN in 1.00 kg of water in a calorimeter caused the temperature of the solution to fall by 30.1 C. Assume the mass of the reactants and products is negligible compared to the mass of the water and that the specific heat of the solution is 4.20 J/g C. Calculate the enthalpy of the reaction in kj/mol: Ba(OH) 2 8H 2O(s) + 2 NH 4SCN(aq) Ba(SCN) 2(aq) + 2 NH 3(aq) + 10 H 2O(l) Activity 201 5 Page 6 of 6