Essay Energy and Calorimeter

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CALORIMETRY

Define: calorimetry, calorimeter, isolated system

State Law of Conservation of Energy.
Energy cannot be created or destroyed, only changed. total heat lost = total heat gained

Law of Thermodynamics
Heat flows from hotter to cooler substances until they reach the same temperature.
Example - metal spoon in cup of hot coffee (The spoon gets hotter and the coffee gets cooler.)

Joule: SI unit of energy

There are three types of calorimeters. The poorest type is the nested styrofoam cups. It is not very accurate because it allows a lot of heat loss to the surroundings. The aluminum can is a better calorimeter. The bomb calorimeter is the best. They all operate on the same principles.

Each calorimeter consists of a reaction chamber in which a known mass of water or solution is present. There is also a thermometer to find the temperature change of this water as the reaction proceeds. Energy is transferred from the reaction to the water or vice versa.

Assumptions made in using a calorimeter to determine a heat of a reaction are that no heat is lost by the system and that the heat transfer occurs only within the calorimeter. Thus we are applying the law that heat lost equals heat gained. We also assume that dilute solutions have the same density and specific heat capacity as water.

If the temperature in the calorimeter is increased, the reaction is exothermic.

If the temperature in the calorimeter is decreased, the reaction is endothermic.

Defining Symbols m = mass (g or kg)
t = change in temperature (oC) c = specific heat (J/goC) q = energy or heat (kJ)
Hm = molar enthalpy or molar heat (kJ/mol)

Example 1: Find the molar heat of combustion of cyclohexane if 5.88 g of cyclohexane are burned in a bomb calorimeter containing 250.0 mL of water at 22.0oC. The final temperature in the calorimeter is 34.0oC.

heatlost = heatgained combustion of = water cyclohexane Ep = Ek cH = Q ncHm = mct

5.88 g x cHm = 0.2500 kg x 4.19 J/goC x (34.0 - 22.0)oC 84.18 g/mol cHm = -180 kJ/mol

Example 2: Find the molar enthalpy of neutralization of H2SO4(aq) if 100.0 mL of 0.50 mol/L H2SO4(aq) solution is mixed with 200.0 mL of NaOH(aq). The initial temperature of the acid and base solution is 20.5oC. The final temperature of the solution is 25.8oC.

heatlost = heatgained neutralization = acid and base chemical change t Ep = Ek nH = Q nnHm = mct nHm = mct n nHm = 300 g x 4.19 J/goC x (25.8 – 20.5)oC 0.50 mol/L x 0.100 L
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