1. CHM 101/102 Laboratory Manual Calorimetry

2. CHM 101/102 Laboratory Manual Calorimetry Background Background  The thermal energy (q) transferred when an object is heated or cooled can be calculated from the following equation: (1) q = m x c x  T heat transferred (in joules) mass of substance (in grams) specific heat capacity of substance (J/gºC) temperature change (in ºC)  T = Tf - Ti

3. CHM 101/102 Laboratory Manual Calorimetry Background Background  When two objects such as hot metal and cold water are brought into contact, the heat lost by the hotter object is equal to the heat gained by the cooler object. (2) - q metal = q water The minus sign is necessary since heat loss is an exothermic process (q is negative) while heat gain is an endothermic process (q is positive).

4. CHM 101/102 Laboratory Manual Calorimetry Background Background  Substituting Eqn (1) into Eqn (2) leads to an equation that allows us to calculate the specific heat capacity of a metal by making a few simple temperature measurements. (3) - (m metal x C metal x  T metal ) = m water x C water x  T water Rearranging Eqn (3) leads to: - m water x C water x  T water m metal x  T metal C metal = (4)

5. CHM 101/102 Laboratory Manual Calorimetry TOC Drop a hot piece of metal into water Heat lost by hot metal = heat gained by cold water

6. CHM 101/102 Laboratory Manual Calorimetry Tips Final temperature of the water and the metal are the same (it reached equilibrium) Final temperature of the water and the metal are the same (it reached equilibrium) If a metal is placed in boiling water, the initial temperature of the metal is 100°C If a metal is placed in boiling water, the initial temperature of the metal is 100°C Water has a special property; 1 : 1 ratio of mass to volume (ie. If I have 10 g of water I have 10 mL of water) Water has a special property; 1 : 1 ratio of mass to volume (ie. If I have 10 g of water I have 10 mL of water) One of the specific heats will always be given to you (either in the problem or in your CRM) One of the specific heats will always be given to you (either in the problem or in your CRM) q of the system = q of the surroundings (just opposite charges) q of the system = q of the surroundings (just opposite charges)

7. CHM 101/102 Laboratory Manual 1. A 10.0 g piece aluminum with an initial temperature of 500.0°C is dropped into a calorimeter with 100.0 g of water. After reaching equilibrium, the temperature of the water is 50.0°C. a. How much heat did the aluminum release? C aluminum = 0.897 J/g°C b. What was the original temperature of the water? C water = 4.184 J/g°C

8. CHM 101/102 Laboratory Manual Problem 2 Calorimetry A 10.0 g piece of aluminum was placed into 150°C hot water bath for a few minutes. It was then placed into a calorimeter containing 200.0 mL of 24.15°C water and the temperature increased to 26.20°C. What is the specific heat of the metal? (remember: Water has a special property of 1:1 mass to volume ratio)