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An experiment designed to measure the specific heats of different materials using a calorimeter. The theory behind the concept of specific heat, the role of a calorimeter in measuring heat transfer, and the steps to perform the experiment. Students will learn how to calculate specific heats using the given data and equations.
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Introductory Mechanics Experimental Laboratory
Goals: Use a calorimeter to measure the change in temperature of its contents. Find the specific heat of different materials.
APPARATUS A calorimeter is a device designed to measure the exchange of heat between two bodies. It works by minimizing the loss of heat from inside the calorimeter to the outside. The calorimeter in this experiment consists of a pot with a narrow opening at the top and a pressure relief nozzle at the side. The pot is designed to use either air or water as buffer, and can be placed over a flame to heat water.
The inner cup of the calorimeter is made of a similar material as the pot and fits in the top hole. The aluminum pot will hold air, and the inner cup will contain water that adjusts to room temperature. The copper pot will hold water and can be placed over a flame to bring the water to boiling. When the inner cup is place inside, any contents will be heated to 100 o^ C with the boiling water.
THEORY Mechanical energy includes the kinetic energy of motion and the potential energy of position. At the microscopic level, atoms and molecules are in constant motion and we can speak of the internal energy of and object. If the substance is an ideal gas, the inter- nal energy is the average kinetic energy of the atoms times the number of atoms. Like kinetic and potential energy, internal energy must be considered for the correct conser- vation of energy.
Temperature is a measure of the internal energy of matter. It is measured in kelvin (K) or oC (0 oC = 273 K). For an ideal gas the internal energy U = (3/2) NkT , where N is the number of atoms, T is the temperature, and k is a constant equal to 1.38 x 10-23^ J/K (Boltzmann’s constant).
When two objects are at different temperatures, the hotter object will lose internal energy and cool off, while the cooler object will gain internal energy and heat up. The energy that is transferred from one object to another because they are at different tem-
peratures is called heat. Like work, heat is a process that changes the energy of on object. Like work it could be measured in J, but is customarily measured calories (1 cal = 4.186 J). The calorie is the amount of heat required to raise the temperature of water by 1 oC.
The amount of heat needed to raise the temperature of other objects depends on the material. The specific heat ( c ) is the measure of how much heat it takes to raise the tem- perature. Mathematically, the heat ( Q ) is related to the mass ( m ) and change in tempera- ture (∆ T ) by EQ 1.
(EQ 1)
Based on the definition of the calorie, water has a specific heat c (^) w = 1.00 cal/g o^ C. Some other specific heats are in Table 1.
Since energy is conserved, the heat gained by any substance equal the heat lost. In our experiment, heat will be gained by an aluminum cup of mass ( m (^) cup ) and water ( m (^) w ) as they increase by a temperature ∆ T 1. The heat will be lost by a metal slug of mass ( m ) as it decreases by a temperature ∆ T 2. EQ 2 represents the balance of heat gained to heat lost.
(EQ 2)
This equation can be solved for the unknown specific heat ( c ).
(EQ 3)
DATA COLLECTION 1. Fill the large copper pot with water. Place the copper pot over a bunser burner flame, and heeat it until boiling. Keep the pot over the flame to maintain the water at the boiling point.
2. Weigh and record the mass of the small aluminum cup ( m (^) cup ). Weigh each of the three metal slugs (aluminum: m (^) Al , copper: m (^) Cu , iron: m (^) Fe ). 3. Fill the aluminum cup about halfway full of water from the sink. Weigh the cup and water, and record the mass difference due to the water ( m (^) w ).
TABLE 1. Specific Heats
Material Specific Heat, c (cal/g oC) Water, cw 1. Aluminum, cAl 0. Iron, cFe 0. Copper, c (^) Cu 0. Lead, cPb 0.
Q = mc ∆ T
m (^) w c (^) w ∆ T 1 + m (^) cup c (^) Al ∆ T 1 = mc ∆ T (^2)
c
( m (^) w c (^) w + m (^) cup c (^) Al ) ∆ T (^1) m ∆ T (^2) =-------------------------------------------------------