PHYS 100 – Calorimetry SAN DIEGO MESA COLLEGE PHYSICS 100 LAB TITLE: Calorimetry Objective: Theory: To determine the specific heat of a metal sample. We have learned that we can explain the motion of objects using Energy. Energy is something that is transferred from one object to another. When the moving cart in lab hit the target cart, some of the kinetic energy of the moving cart was transferred to the target cart. In these interactions, objects can be either energy givers or energy receivers. Energy givers lose energy and energy receivers gain energy. In all of our examples so far, we have been concentrating on explaining why things move. Our evidence has been the change in speed in of objects. We interpret changes in speed as changes in Kinetic Energy. If an object speeds up, the Kinetic Energy increases and the object must be receiving energy. If the object slows down, the Kinetic Energy decreases and the object must be giving energy. We are going to change our focus slightly to a different interpretation of Kinetic Energy. As we saw in class, Temperature is another measure of Kinetic Energy. When the temperature of an object increases, the kinetic energy of the particles that make it up increase as well. The Work/Energy theorem still applies here. When an object increases in temperature, its kinetic energy increases and it is receiving energy. When an object decreases in temperature, its kinetic energy decreases and it is giving away energy. The energy that an object gives or receives in this case is called Heat. Heat has the symbol Q. When a heat transfer causes a change in temperature, the heat can be calculated as: Q = cm∆T. This equation is saying that the energy depends on how much you want to change the temperature of an object (∆T) and how much material that you have (m – mass). It also depends on the material itself and how easy it is to change its temperature. We call this quantity, specific heat or c. The specific heat is measured in J/kg 0C or sometimes cal/g℃. The specific heat is the amount of energy it takes to raise 1 gram of a substance by 1 oC. The unit of energy transferred through heat is the calorie. One calorie is the amount of heat required to raise one gram of water by 1centigrade degree. This experiment makes use of a Calorimeter. A calorimeter is a very insulated cup in which you can put materials of different temperatures and allow them to mix and come to equilibrium. The insulation allows for energy transfers between the two materials, but not to the outside environment. The calorimeter used in the laboratory is a Styrofoam cup. Many of the insulated water bottles that people use to keep their water cold would make a nice calorimeter. You may have noticed that if you fill your water bottle with water and ice cubes, the ice takes a long time to melt. Mostly, heat is transferred from the water to the ice cubes. If the water is very cold to begin with, the water/ice will come to equilibrium with some ice remaining. For our lab, we will be using a calorimeter with some water in it at a known temperature. Different substances of known temperature will be placed in the calorimeter. The system will come to equilibrium. We will assume that there is no heat transferred to the environment and from this we can calculate heat loss and gains. Online Site: https://media.pearsoncmg.com/bc/bc_0media_chem/chem_sim/calorimetry/Calor.php 1 PHYS 100 – Calorimetry Set up The apparatus in the animation has two parts. First, the Sample Cup is shown on the left. You can put liquids, solids or solutions into the beaker and choose the mass and Temperature of the substance. On the right is the Calorimeter. The Calorimeter is filled with a certain mass of liquid (we are going to stick with water) at a certain temperature. The lid is closed and the liquid is mixed. Sample Calorimeter Procedure: The specific heat of a substance is measured by the method of mixtures. A known mass of the substance at a high temperature is placed in a known mass of water at low temperature. The mixture is allowed to reach thermal equilibrium. The specific heat is then calculated using energy conservation and neglecting the loss of heat during transfer. Follow the directions in the Lab write up carefully. 2 PHYS 100 – Calorimetry Calorimetry Name________________________________________________ PART 1-INTRODUCTION  Start by clicking on the “liquid” tab and choose “Water H2O”. Use the Mass sliders to select 100 g and the Temperature slider to select 500C. Click on the “Next” button.  Choose liquids again to put 100 g of Water at 200C into the Calorimeter. Click on the “Next” button.  Before you Run Experiment, click on the “Show graph view” option. Then click on Start. The animation is mixing together the Cold and Warm Water in the calorimeter. The graph is showing you the temperature of the mixture over time. Eventually, the mixture comes to equilibrium when the heat transfer processes stop and all of the water is at the same temperature. That temperature is displayed on the Temperature probe on the calorimeter.  Which water is the energy giver in this case, the Cold H2O or the Hot H2O? How do you know?  Which water is the energy giver in this case, the Cold H2O or the Hot H2O? How do you know?  Use the information from the interactive to fill in the table and calculate the Heat for each substance. The Specific Heat for water is 1.00 cal/g C0. Giver Receiver 1.00 1.00 Mass (m) in g ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 Specific Heat (cal/g C0) 𝑄 = 𝑐𝑚∆𝑇  Was the Heat lost by the Energy giver equal to the Heat gained by the energy receiver? In the “real world” what could happen in the experiment so that the Heat lost is NOT equal to the Heat gained? 3 PHYS 100 – Calorimetry PART 2: SPECIFIC HEAT OF COPPER Click on the “Reset” button. 1. Click on the “solid” tab and choose “Copper”. Use the Mass sliders to select 20 g and the Temperature slider to select 2000C. Click on the “Next” button. 2. Choose liquids again to put 100 g of Water at 200C into the Calorimeter. Click on the “Next” button. 3. Before you Run Experiment, click on the “Show graph view” option. Then click on Start.  Which water is the energy giver in this case, the H2O or the Copper? How do you know?  Assume that the Heat lost by the energy giver is equal to the Heat gained by the receiver. Fill in the information that you used in the interactive and that water has a specific heat of 1.00 cal/g C0 and calculate the specific heat of the copper. Giver Receiver Mass (m) in g ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 Specific Heat (cal/g C0) 𝑄 = 𝑐𝑚∆𝑇  The actual specific heat of copper is 0.0923 cal/g C0. Find the Percent Difference using the following equation: %𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑐𝑒𝑥𝑝 −0.0923 0.0923 𝑥100.  What could account for any differences? 4
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