Calorimetry

Calorimetry -- the precise measurement of the heat flow into or out of a system for chemical and physical processes. In calorimetry, the heat released by the system is equal to the heat absorbed by its surroundings. Conversely, the heat absorbed by a system is equal to the heat released by its surroundings. What law does that sound like?

Review We talked yesterday about specific heat and the equation associated with it. The thermal energy (q) transferred when an object is heated or cooled can be calculated from the following equation: q = m x c x DT heat transferred in joules (J) Change in temperature ∆T = Tfinal – Tinitial in 0C Mass of substance in grams (g) Specific heat in J/g0C

Conservation of Energy Since energy cannot be created or destroyed; 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. - qmetal = qwater Remember that the negative sign is there to indicate what direction the heat is moving. Since the metal is releasing heat the q is negative (exothermic); the water gaining heat causes q to be positive (endothermic).

(mmetal x Cmetal x DTmetal) = mwater x Cwater x DTwater Combining Equations We can now take the two equations together to use all the information about each substance. (mmetal x Cmetal x DTmetal) = mwater x Cwater x DTwater This equation can be rearranged to solve for any of the variables. For example, if we wanted the specific heat of the metal: - mwater x Cwater x DTwater Cmetal = mmetal x DTmetal

Calorimeter The equipment used to measure the heat flow is a calorimeter. Its insulated design is meant to minimize the loss of heat while measuring the transfer. The metal would be placed inside the calorimeter, which has water in it, and the temperatures measured. Heat lost by metal = heat gained by water

Calorimetry Tips 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 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) q of the system = q of the surroundings (just opposite charges)

Practice 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. How much heat did the aluminum release? Caluminum = 0.897 J/g°C What was the original temperature of the water? Cwater = 4.184 J/g°C q = m c DT m = 10.0g C = 0.897 J/gC ∆T = 450.00C ∆Twater = mAlCAl∆TAl/mwaterCwater ∆Twater = 9.660C ∆T = Tfinal – Tinitial 9.66 = 50.0 –T T = 50.0-9.66 = 40.30C q = (10.0)(0.897)(450) 4036.5 J 4040 J

More Practice - mwater x Cwater x DTwater Cmetal = mmetal x DTmetal 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) - mwater x Cwater x DTwater Cmetal = mmetal x DTmetal CAl = -(200.0)(4.184)(2.05)/(10.0)(123.8) = -1.38J/g0C