Calorimetry

Calorimetry: A laboratory technique that determines the amount of heat transferred in a chemical or physical process

Calorimeter The instrument used to minimize the heat lost to the surroundings

This is what a calorimeter looks like (cheap version): Two Styrofoam Cups Thermometer

How a calorimetry experiment works: Water is placed in the calorimeter and its initial temperature is taken A second substance is heated and its initial temperature is recorded

Both masses (the water and the substance) are recorded The “substance” is placed in the water and the lid is quickly placed on The experimenter records the temperature change in the water until the temperature is constant for a number of readings The final temperature is recorded. This is the final temperature of the water and the “substance”

Heat lost by one is gained by the other (because energy is conserved) Calculations: When two substances are added to each other and allowed to come to one temperature, that is called dynamic equilibrium. Heat lost by one is gained by the other (because energy is conserved)

qlost + qgained = 0

Example of problems: A 0.45 gram piece of copper (C=0.385 J/g(oC)) at a temperature of 87.0oC is placed in a calorimeter. The calorimeter contained water (C=4.184 J/g(oC)) at a temperature of 23.0oC. The solution came to equilibrium at 24.8oC, what was the mass of water in the calorimeter?

Givens: Copper m = 0.45 grams Cu CCu = 0.385 J/g(oC) Tinitial = 87.0oC (metal) Tfinal = 24.8oC Water CH2O = 4.184 J/g(oC) Tinitial = 23.0 oC (water)

Copper q = mC T q = (0.45)(0.385)(24.8-87.0) q = - 10.776 J That is the heat lost (- 10.776 J) so the water must have gained 10.776 J Therefore qwater = 10.776 J

Water q = mC T 10.776 = m(4.184)(24.8-23.0) m = 1.4 grams H2O Since we knew q for copper, we could use that number (with the opposite sign) to get q for water since heat lost by one is gained by the other.

Another example: A piece of titanium metal with a mass of 20.8 g is heated in boiling water to 99.5oC and then dropped into a calorimeter containing 75.0 g water (C= 4.184 J/g(oC)) at 21.7oC. The final temperature in the calorimeter is 24.3oC. What is the specific heat of Titanium?

Givens: Titanium m = 20.8 g Ti = 99.5oC Tf = 24.3oC Water m = 75.0 g C = 4.184 J/g(oC)

Water q = mC T q = 75.0(4.184)(24.3-21.7) q = 815.88 J If water is gaining 815.88 J, then the titanium has lost 815.88 J so make it negative and use it for qtitanium.

Titanium q = mCT -815.88 = 20.8(C)(24.3-99.5) -815.88 = -1564.16(C) C = 0.522 J/g(oC)