Calorimeters and Calorimetry

Calorimeters and CalorimetryCalorimetry is the skill associated with determining the changes in muscle of a system by measuring rod the take fire deepend with the surroundings. Now that sounds very textbooky but in this last part of Lesson 2, we be going to try to make some meaning of this interpretation of calorimetry. In physics class (and for some, in chemistry class), calorimetry labs atomic number 18 frequently performed in ordinance to determine the heat of reaction or the heat of coalescency or the heat of dissolution or even the specific heat capacity of a metal. These types of labs are rather popular beca example the equipment is relatively inexpensive and the measurements are usually straightforward. In such labs, a calorimeter is employ. A calorimeter is a device used to measure the quantity of heat transferred to or from an object. Most students likely do not remember utilize such a fancy piece of equipment known as a calorimeter. Fear not the reason for the l ack of repositing is not a sign of early Alzheimers. Rather, it is because the calorimeter used in high school science labs is more commonly referred to as a Styrofoam instill. It is a burnt umber instill calorimeter usually filled with water. The more sophisticated cases include a lid on the cup with an inserted thermometer and maybe even a stirrer.Coffee loving cup CalorimetrySo how fecal matter such simple equipment be used to measure the quantity of heat gained or lost by a system? We conduct learned on the previous page, that water will change its temperature when it gains or loses power. And in fact, the quantity of talent gained or lost is given by the parQ = mwaterCwaterTwaterwhere Cwater is 4.18 J/g/C. So if the mass of water and the temperature change of the water in the coffee cup calorimeter can be measured, the quantity of energy gained or lost by the water can be calculated.The assumption behind the science of calorimetry is that the energy gained or lost by the water is equal to the energy lost or gained by the object under study. So if an attempt is being made to determine the specific heat of fusion of ice using a coffee cup calorimeter, then the assumption is that the energy gained by the ice when melting is equal to the energy lost by the surrounding water. It is fictive that there is a heat exchange amid the iceand the water in the cup and that no other objects are involved in the heat exchanged. This tilt could be placed in equation form asQice = Qsurroundings = -QcalorimeterThe role of the Styrofoam in a coffee cup calorimeter is that it reduces the amount of heat exchange between the water in the coffee cup and the surrounding air. The value of a lid on the coffee cup is that it also reduces the amount of heat exchange between the water and the surrounding air. The more that these other heat exchanges are reduced, the more true that the above mathematical equation will be. Any erroneous belief analysis of a calorimetry ex amine must take into consideration the flow of heat from system to calorimeter to other parts of the surroundings. And any design of a calorimeter experiment must give attention to reducing the exchanges of heat between the calorimeter contents and the surroundings.Bomb CalorimetryThe coffee cup calorimeters used in high school science labs provides students with a worthwhile exercise in calorimetry. But at the professional level, a cheap Styrofoam cup and a thermometer isnt going to assist a commercial message food manufacturer in determining the Calorie content of their products. For situations in which exactness and accuracy is at stake, a more expensive calorimeter is needed. Chemists often use a device known as a bomb calorimeter to measure the heat exchanges associated with chemical reactions, especially combustion reactions. Having little to nothing to do with bombs of the host variety, a bomb calorimeter includes a reaction chamber where the reaction (usually a combustion reaction) takes place. The reaction chamber is a strong vessel that can apply the intense pressure of heated gases with exploding. The chamber is ordinaryly filled with mostly oxygen gas and the fuel. An electrical circuit is wired into the chamber in order to electrically ignite the contents in order to perform a study of the heat released upon combustion. The reaction chamber is surrounded by a cover of water with a thermometer inserted. The heat released from the chamber warms the water-filled jacket, allowing a scientist to determine the quantity of energy released by the reaction.Source Wikimedia Commons thanks to Lisdavid89.Solving Calorimetry bothersNow lets pure tone at a few examples of how a coffee cup calorimeter can be used as a tool to answer some typical lab questions. The next three examples are all based on laboratory experiments involving calorimetry.Example Problem 1A physics class has been charge the task of determining an experimental value for the heat of fusion of ice. Anna Litical and Noah Formula dry and mass out 25.8-gram of ice and place it into a coffee cup with 100.0 g of water at 35.4C. They place a lid on the coffee cup and insert a thermometer. After several minutes, the ice has all melted and the water temperature has lowered to 18.1C. What is their experimental value for the specific heat of fusion of ice?The basis for the solution to this problem is the actualization that the quantity of energy lost by the water when cooling is equal to the quantity of energy required to melt the ice. In equation form, this could be verbalize asQice = -Qcalorimeter(The detrimental sign indicates that the ice is gaining energy and the water in the calorimeter is losing energy.) Here the calorimeter (as in the Qcalorimeterterm) is considered to be the water in the coffee cup. Since the mass of this water and its temperature change are known, the value of Qcalorimeter can be determined.Qcalorimeter = mCTQcalorimeter = (100.0 g)(4.18 J/g /C)(18.1C 35.4C)Qcalorimeter = -7231.4 JThe negative sign indicates that the water lost energy. The assumption is that this energy lost by the water is equal to the quantity of energy gained by the ice. So Qice = +7231.4 J. (The positive sign indicates an energy gain.) This value can be used with the equation from the previous page to determine the heat of fusion of the ice.Qice = miceHfusion-ice+7231.4 J = (25.8 g)Hfusion-iceHfusion-ice = (+7231.4 J)/(25.8 g)Hfusion-ice = 280.28 J/gHfusion-ice = 2.80102 J/g (rounded to two significant figures)Example Problem 2A chemistry student dissolves 4.51 grams of sodium hydrated oxide in 100.0 mL of water at 19.5C (in a calorimeter cup). As the sodium hydroxide dissolves, the temperature of the surrounding water increases to 31.7C. modulate the heat of solution of the sodium hydroxide in J/g.Once more, the solution to this problem is based on the recognition that the quantity of energy released when sodium hydroxide dissolves is equal to t he quantity of energy absorbed by the water in the calorimeter. In equation form, this could be stated asQNaOH dissolving = -Qcalorimeter(The negative sign indicates that the NaOH is losing energy and the water in the calorimeter is gaining energy.) Since the mass and temperature change of the water have been measured, the energy gained by the water (calorimeter) can be determined.Qcalorimeter = mCTQcalorimeter = (100.0 g)(4.18 J/g/C)(31.7C 19.5C)Qcalorimeter = 5099.6 JThe assumption is that this energy gained by the water is equal to the quantity of energy released by the sodium hydroxide when dissolving. So QNaOH-dissolving = -5099.6 J. (The negative sign indicates an energy lost.) This quantity is the amount of heat released when dissolving 4.51 grams of the sodium hydroxide. When the heat of solution is determined on a per gram basis, this 5099.6 J of energy must be divided by the mass of sodium hydroxide that is being dissolved.Hsolution = QNaOH-dissolving / mNaOHHsolution = ( -5099.6 J) / (4.51 g)Hsolution = -1130.7 J/gHsolution = -1.13103 J/g (rounded to three significant figures)Example Problem 3A large paraffin candle has a mass of 96.83 gram. A metal cup with 100.0 mL of water at 16.2C absorbs the heat from the burning candle and increases its temperature to 35.7C. Once the burning is ceased, the temperature of the water was 35.7C and the paraffin had a mass of 96.14 gram. Determine the heat of combustion of paraffin in kJ/gram. GIVEN density of water = 1.0 g/mL.As is always the case, calorimetry is based on the assumption that all the heat lost by the system is gained by the surroundings. It is assumed that the surroundings is the water that undergoes the temperature change. In equation form, it could be stated that