Thermochemistry Lecture 1.

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Presentation transcript:

Thermochemistry Lecture 1

Thermochemistry Study of heat changes of chemical reactions and phase changes System: specific part of the universe that contains the reaction or process you are studying Surrounding: everything in universe except the system

Energy Ability to do work or produce heat 2 Forms Kinetic Potential Law of Conservation of Energy Energy can be converted from one form to another in a reaction, but it is neither created nor destroyed

Heat Symbol : q Energy that is in the process of flowing from a warmer object to cooler object Measured in Joules (J)

Specific Heat Amount of heat required to raise the temperature of one gram of that substance by 1°C. Measure of how efficiently that substance absorbs heat (directly proportional) Symbol : c Water = 4.184 J/g*°C Water has a high specific heat allowing it to absorb or release large quantities of energy

Heat Absorbed or Released by objects The heat absorbed or released by a substance during a change in temperature depends on specific heat, change in temperature and the mass of the substance. q = c*m*ΔT ΔT = final T – initial T

Practice Problem If the temperature of a 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed by the ethanol? Specific heat of ethanol: 2.48 J/g*°C 4.52*10^3 Joules

Measuring heat Calorimetry Insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process Can be used to calculate the specific heat of an unknown

Practice Question If a 50 g unknown metal with a temperature of 115.0°C, is placed in 125 g of 25.6°C water in a foam cup calorimeter and the metal causes the water to raise it’s temperature to 29.3°C, what is the specific heat of the metal? Part 1: solve the heat the water gained Part 2: heat gained by water = heat lost by metal Part 3: Using the change in temperature of the metal = final T of water – initial T of metal Part 1: q=mcT Q = 4.184 * 125 * (29.3-25.5) = 1900 J Part 2 1900 J = q(metal) = mcT Part 3 1900 J = 50 g * c * (115-29.3) C= 0.44 J/g*°C

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