Specific Specific HeatHeat Capacity Capacity

deals with heat changes deals with heat changes that occur during chemical that occur during chemical reactionsreactions

Heat – transfer of energy Heat – transfer of energy from one object to anotherfrom one object to another

ThermochemistryThermochemistry

EndoEndothermicthermic – heat flows – heat flows into into the systemthe system the test tube gets cold – heat is the test tube gets cold – heat is absorbedabsorbed

ExoExothermicthermic – heat flows – heat flows outout of the system of the system the test tube gets hot – the test tube gets hot – heat is given offheat is given off

““Burning” caloriesBurning” calories 1 1 ccalorie = quantity of heat needed alorie = quantity of heat needed

to raise the temperature of 1 gram to raise the temperature of 1 gram of water 1 of water 1 ooCC

CCalorie = dietary calorie = 1,000 alorie = dietary calorie = 1,000 calories = 1 kilocaloriecalories = 1 kilocalorie

1 Joule (metric unit) = 0.239 1 Joule (metric unit) = 0.239 caloriescalories

4.184 Joules = 1 calorie4.184 Joules = 1 calorie 10 g of sugar =10 g of sugar =

41 Calories of heat or41 Calories of heat or 41 kilocalories of heat41 kilocalories of heat

Heat CapacityHeat Capacity the amount of heat needed to increase the the amount of heat needed to increase the

temperature of an object by 1 temperature of an object by 1 ooCCdepends on the depends on the massmass and chemical and chemical

compositioncomposition of the matter of the matter

250 gram steel cylinder

50 gram steel cylinder

greater heat capacity

wooden block

iron block

greater heat capacity

SpecificSpecific Heat Capacity Heat Capacity

The amount of heat it takes to raise the The amount of heat it takes to raise the temp. of 1 gram of a substance 1 temp. of 1 gram of a substance 1 ooCC

calorimetrycalorimetry = the measurement of heat = the measurement of heat changeschanges

Some specific heat capacitiesSome specific heat capacities

Substance J/(gXoC)

water 4.18

ice 2.1

aluminum 0.90

glass 0.50

iron 0.46

silver 0.24

Calculation of the specific heat of a Calculation of the specific heat of a substancesubstance

S = specific heatS = specific heatqq = heat = heatm = massm = mass∆∆T = change in temperature = TT = change in temperature = T22 – T – T11

q = m S ∆T q = m S ∆T or q = m C ∆T or q = m C ∆T

Some problems!Some problems!

The temperature of a piece of copper with a mass of The temperature of a piece of copper with a mass of 95.4 g increases from 25.0 95.4 g increases from 25.0 ooC to 48.0 C to 48.0 ooC when the metal C when the metal absorbs 849 J of heat. What is the specific heat of absorbs 849 J of heat. What is the specific heat of copper?copper?

m = 95.4 g

∆T = (48.0 oC – 25.0 oC) = 23.0 oC

q = 849 J

S = ___q___

m ∆T

S = ______849J_______

95.4 g X 23.0 oC

= 0.387 J/(g oC)

When 435 J of heat is added to 3.4 g of olive oil When 435 J of heat is added to 3.4 g of olive oil at 21 at 21 ooC, the temperature increases to 85 C, the temperature increases to 85 ooC. C. What is the specific heat of olive oil?What is the specific heat of olive oil?

S = ____435 J_____

3.4 g X 64 oC

S = 2.0 J/(gX oC)