Exothermic and Endothermic changes

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 Energy is transferred ◦ Exothermic – heat is released  Heat exits ◦ Endothermic – heat is required or absorbed by reaction  Measure energy in Joules.
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Presentation transcript:

Exothermic and Endothermic changes the change in heat of the water tells us about the reaction of the chemicals. An increase in water temperature indicates that the chemicals released energy when they reacted. This is called an “exothermic” reaction. In an “endothermic” reaction, water temperature decreases as the chemicals absorb energy. We will see that heat is measured in Joules (J) or kiloJoules (kJ).

Heat is the energy which flows between objects w/ 2 different temperatures. Heat always flows from high temp to low temp.

Thermal energy is the total amount of energy in an object. Temperature: is a measure of the average Kenetic Energy of a substance’s particles (molecules/atoms). All atoms vibrate High Temp-High KE-fast motion Low Temp-Low KE-slow motion

Difference between Thermal and Total Energy Thermal is the measure of the actual amount of energy in a substance (KE) Temp is just the average (KE) of all the particles of a substance but you don’t know how many particles there are ex: pool vs. cup Heat is the transfer of thermal energy from one substance to another. Heat is measured in Joules/Kilojoules (joules x 1000).

Specific heat capacity The heat needed to  the temperature of 1 g of a substance by 1 C. Symbol: c, units: J/(gC). Specific heat of water is 4.18j/g C 1g of H2O at 10 C -- 1g of H2O at 11 C How many joules of heat did it take to raise the temp by one degree? Each substance has it’s own unique specific heat 5/27/2019

Heat Calculations To determine the amount of heat a substance produces or absorbs we often use q = cmT q: heat in J, c: specific heat capacity in J/(gC), m: mass in g, T: temperature change in C, T= TF-Ti Sample problem: (c of water = 4.18 J/gC) When 12 g of a food was burned in a calorimeter, the 100 mL of water in the calorimeter changed from 20C to 33C. Calculate the heat released. q=cmT = 4.18 J/(gC) x 100 g x 13C = 5.4 kJ

More practice 5 g of copper was heated from 20C to 80C. How much energy was used to heat the Cu? Remember!! There is no change in temperature when there is a phase change, this equation does not apply!! The heat added or removed during a phase change does not change the temp. q=cmT = 0.38 J/(gC) x 5 g x 60 C = 114 J