4. Remove the object from the calorimeter. dry off the sample and determine its mass (in grams). WE QMECT BACK H1111? THE HQT ATERl!!!!I...this may affect other group's measurements and may break the beaker if dropped in. Place the object M to the beaker you got it from. 5. Obtain a new 100 ml of cold water and repeat steps 2-5 with each new sample. Be sure to re-measure the initial temperature of both the water and new object. ** You do not have to try and find the specific heat of water. You are to assume that 5\Lab: Calorimetry mm Eval c To be able to describe the meaning of the specific heat of a material and represent it with the correct units. c To be able to describe calorimetry as a restatement of conservation of energy, and describe the storage and transfer of energy (heat) at different times e To be able to solve calorimetry problems. c To be able to understand that all measurements have some error and be able to identify sources of error. Emma: In this lab. you will be calculating the specific heat of several materials. Specific heat is the amount of energy needed to raise the temperature of 1 gram of an object by 1C. materials that have a high specific heat require more energy to heat up, and release more energy when they cool down, than do materials with a low specific heat. Conservation of Energy states that energy cannot be created or destroyed. Previously, we had said that energy is often \"lost\" to friction in a moving object. Thermodynamics allows us to take into consideration the amount of energy transferred to heat by friction. The Second Law of Thermodynamics states that energy flows from areas of high temperature to areas of low temperature. The First Law of Thermodynamics takes this another step and states that WM): a hat abjecj will be eaml to the amnt cf heat gained by a cold object. In this lab you will have to relate the heat of a astem before and interaction to the heat of the system afterward. Remember that heat is related to specific heat by the equation, (1) Q = chT where c is the specific heat. The specific heat of water is 1 cal/gyc or 4.136 Jig-\"C. *** A relationship (equation) that says the amount of heat lost by a hot object will be equal to the amount of heat 991W using equation (1) is shown in equation (2). This is just another form of conservation of energy. (2) Q5051 by water + anined by object = O ** Q...\" um will turn out negative since its temperature is decreasing. murals: calorimeter, thermometer, sample materials, graduated cylinder, electric balance, hot plate, tongs, beakers. Procedure: (You will need to test each of your materials separately.) 1. Place 100 mL of cold water in an empty calorimeter. Measure the temperature of the water. 2. Obtain a sample of one of the materials and measure the temperature of the water it is in. This is also 1h; gamma. Make sure you bring your calorimeter with you and transfer the object quickly. 3. Put the sample directly into your calorimeter, insert the thermometer. and swirl gently. Record the final temperature of the water and object mixture when the thermometer reaches its highest temperature. The water and object should now be at the same temperature {thermal equilibrium)