Calorimetry and Specific Heat Measuring Heat Transfer in Metals

1- a)Introduction Calorimetry is the science of measuring heat. We will use a calorimeter to measure the amount of heat transferred from the two different metals to water. By knowing the value of the specific heat of water we will then be able to calculate the specific heat of the both metals. Our metals are Copper, and Aluminum. Also when a substance is heated, the motion of its individual particles increases, resulting in an increase in temperature. The more heat that is added per gram of substance, the greater the temperature change. The relationship between the heat added, the mass of a substance, and the temperature change it undergoes is known as specific heat. Specific Heat ｘ Temperature change in = (Energy change in calories.) /Mass in grams Celsius). Specific heat is defined as the amount of energy necessary to produce a temperature change of 1°C per gram of substance. The specific heats of different substances vary, and therefore this quantity may be useful in identifying an unknown. b) Objective The purpose of this lab was to find the total heat loss of the water when placed in the calorimeter. Changes in heat can be determined by measuring changes in temperature which is what was measured in this lab. The heat gained depends on the mass of the substance in grams, the specific heat of the substance, and the change in temperature of the substance. In this lab we calculate all of these things to find the heat capacity of the Calorimeter. 2- a) Theory The Specific heat of a substance is an intensive property that relates the temperature and heat of an object. It is defined as the quantity of heat required to raise 1 gram of a substance by 1 and has units of (J/g, cal/g, or J/mol). The relationship between heat (q), specific heat (s), mass (m) and a temperature change (∆t or tf − ti) can be expressed mathematically as: q = m · s · ∆t (1) The change in temperature of a substance is often expressed using the greek letter delta (∆) “which means the change in”. This can also be expressed as the final temperature (tf ) minus the initial (ti) of an object. Knowing 3 of the variables in the equation above allows us to solve for the fourth variable. A fact we will make use of in this experiment. Based on our definitions of specific heat and joules/calories, the specific heat of water is exactly 1.00 cal/g. Knowing this value we will be able to determine the specific heat of a known metal and an unknown metal. We will do so by heating the metal to a known temperature and then placing it in contact with water at a known temperature. The heat from the metal will be transferred to the water raising the temperature. Using the final measured temperature of the water we will be able to determine the amount of heat transferred from the metal to the water and thus calculate the specific heat of the metal. b) Equations Q= (M)(S)( ΔT¿ M m ×S m ×(t m −t f )= M w ×S w ×(t f −t o)+M c ×S c (t f −t o ) ⇒ M w ×S w ×(t f −t o)+ M c ×S c (t f −t o ) S m=( ) M m ×(t m −t f ) c) Diagrams 3- Experiment a) Equipment 1.Lab Calorimeter, 2.Thermometer, 3.Hot water, 4.room temperature water, 5.beam balance, 6.Beaker. b) Experimental; diagrams c) Procedure Weigh the Calorimeter and record the weight in your lab notebook. Add 50-70 mL of cold water from the tap into the Calorimeter. Weigh the Calorimeter again with the cold water in it and record it. Place thermometer into Calorimeter to get initial temperature of the cold water. This will also serve as the initial temperature of the Calorimeter. Record this in your lab notebook. Use the Weight of the Calorimeter minus the weight of the Calorimeter with the cold water to find the mass of the cold water alone and record this. Fill boier ⅔. After weighing everything required, place metal shots cup in boiler and heat it. Record initial temp of water in calorimeter, transfer metal to water and record the final temp. Calculate specific heat, and determine the % error of standard and calculated specific heat. 4- Results a) Data table Copper Aluminum Mass of empty boiler cup 64.5 g 64.5 g Mass of Boiler cup plus metal shots 443.9 g 188.8 g Mass of Metal Shots ( M m ¿ 379.4 g 124.3 g Mass of calorimeter inner cup + stirrer ( M c ) 59.8 g 59.8 g 213.3 g 211.5 g Mass of calorimeter inner cup + stirrer + water ( M c +M w ) Mass of water ( M w ) 153.5 g 151.7 g t (¿¿m) ¿ 97 C 96 C Initial Temp. of water and cup ( t o ¿ 23 C 23 C 35 C 33 C Change in Temp. of metal shots ( t m−t f ) 62 C 63 C Temp. change of water and cup ( t f −t o ) 12 C 10 C Initial Temp. of metal shots Equilibrium Temp. t (¿¿f ) ¿ Copper 1) energy lost by copper shots = 23,522 2) energy gained by cup + stirrer + water = 1,992 3) Specific heat of copper = 0.085 Aluminum 1) energy lost by Aluminum shots = 7,812 2) energy gained by cup + stirrer + water = 1,664 3) Specific heat of Aluminum = 0.21 b) Graph No Graph. c) Error Calculations, Accuracy In my calculations, my result is not that different from the standard specific heat, and I think that difference is because of transfer of energy into some other types so energy, so our specific heat is a little off. But in my opinion I think our results don’t have much difference. In copper our error is 8% and in Aluminum it is 4%. 5- Discussion a) Possible sources of error Heat can be lost through the heating chamber , also heat can be lost through the thermometer. Heat lost through an inefficiency in transfer from the heating element to the water. Noncalibrated thermometer, or general inaccuracies in the temperature measurement (even if the thermometer is calibrated). Non-uniform heating of the water measuring the temperature in a 'hot' or 'cold' spot. If stirring the water to prevent non-uniform heating, the addition of energy to the water, from the stirring device. Aslo taking more time to transfer metal to water will cause loss of energy and our result will not be precise. The human error is common, like doing calculations someone can get wrong answer. b) Observations I observed that, if we solve and get the specific heat of an substance, we can easily determine what is the actual substance is this, by comparing to the tables. I also observe that the experiment needs a lot of attention, and accuracy to read thermometer. If reading thermometer is not correct, we messed up. Also I think time to transfer should be less, and it has to be done fast, so it don’t lose more energy in air. c) Suggestions I will suggest that, keep focus on the variables, which is what. Which symbol belong to what name. I also suggest you that read thermometer carefully, and be attentive when it goes to above 95 C, keep an eye on it. I will also suggest that, when the water is boiling in the boiler, keep distance, because water bubbles are coming out, adn or very hot. And while doing calculations, make no mistakes. 6- Conclusion After the experiment, I can conclude that a metal has a property that can absorb and emit heat. This property, which is the specific heat, depends on most of the metals like the Aluminum and Copper. Also, the specific heat of each metal can only be measured in such temperature-sensitive materials because of its very small actual values that cannot be easily measured on other metals. In performing the experiment we should consider some factors in order to not repeat the experiment. First is the knowledge that Aluminum conducts more heat than Copper. Secondly, Copper metals and other metals with low specific heat are usually the ones that emit more heat. And lastly, in both experiments, we should focus in measuring the initial and final temperatures. In our experiment I found that copper has specific heat = 0.085 ; meanwhile, aluminum has = 0.21. These are the numbers which I found by doing experiment and calculations in the end, but the standard specific heat of both is slightly different. Copper standard specific heat is 0.085 ; and aluminum has = 0.22. Error in copper is 8% and in Aluminum is 4%. Questions for discussion 1) I think the energy is transferred by conduction, because both water, and metal shots were in contact, so the water conduct heat from the metal shots. 2) Energy lost is much bigger, because all the energy is not gained by water, but most of the energy is lost by metal shots. Because energy transferred to other forms of energy, so all the lost energy is not gained by the water, and this is expected, according to the law of energy. 3) Experimental errors are we transferred the energy in to the atmosphere, and to the objects nearby to the water, as calorimeter outside cup, and stirrer and calorimeter. A lot of energy which is lost by metal shots but some of energy is gained by the water, if we have used better equipment and in also in heat chamber that will be better. Because heat is also transferred to other types of energy forms. When we transfer the metal shots to the water most of the energy is lost at that time.