Energy and energy transformations. First Law of Thermodynamics Energy is never created nor destroyed Energy can change forms, but the quantity is always.

THERMODYNAMICS

Energy and energy transformations

First Law of Thermodynamics Energy is never created nor destroyed

Energy can change forms, but the quantity is always constant.

Second Law of Thermodynamics

The Entropy of the Universe is always increasingEntropy= disorder

Third Law of Thermodynamics

The entropy of an ideal solid at zero Kelvin is zeroAll molecular motion stops at 0 K

Endothermic Reactions Energy is used to

begin a reaction Products have

higher energy than reactants

Absorbs heat from surroundings Ice meltingWater evaporating

Exothermic Reactions Gives off energy

during a reaction Reactants have

more energy than products

Gives off heat Ice freezingWater condensing

Heat vs Temperature

Temperature - measure of average KE

Heat - measure of energy transfer

Temp change (∆T) depends on: amount of heat

transferred (q) mass of object (m) specific heat of the object

(C) Video link: http://

www.youtube.com/watch?v=wTi3Hn09OBs&safety_mode=true&persist_safety_mode=1

Measuring Heat q = m C ∆T

q = heat○ measured in joules (J)

or kilojoules (kJ) ○ 1000J = 1kJ

m = mass; measured in grams (g)

∆ T = change in temp○ measured in Celsius

(°C) or Kelvin (K)C = specific heat

○ units are J/g°C or J/gK

Endo/exothermic -q = release heat (exothermic) +q= absorb heat (endothermic)

What amount of heat is needed to increase the temp of 10 g of Hg by 5C? (specific heat of Hg is 0.139 J/gC)

If 68,000J of heat are added to 25g of H2O (specific heat of water is 4.184 J/gC), what is the change in temperature in Celsius?

Calorimetry Calorimetry

uses a closed system (calorimeter) to determine the energy change or specific heat of an unknown substance

Calorimeter- an insulated device usually filled with water or a substance with a known specific heat

1st law - conservation of energy When applied to a closed system, any

energy that is lost by one substance is gained by the other

qsubstance1 = - qsubstance2

since q = m C ∆ T…

m1 C1 ∆ T1 = - m2 C2 ∆T2

Ex: A 55.8g piece of unknown metal at 180°C is placed into 100.0g of water that began at 25°C. The final temperature of both was 26.8°C. If the specific heat of water is 4.184 J/g°C, calculate the specific heat of the metal.