Thermochemistry:

The study of heat changes that occur during chemical

reactions and physical changes of state

Energy

• Heat (q) = – Form of energy – always flows from a warmer object to a

cooler object

• Law of Conservation of Energy – Energy is neither created nor destroyed

Heat (q)

• The SI unit is the joule• 1 Cal (food) = 1000 cal (science) = 1

kcal• 1 cal = 4.186 J

Specific Heat (C)• Specific Heat (C) – the amount of heat

required to raise 1 gram of a substance by 1C

• Specific heat is an intensive property,• Every substance has its own specific

heat• Ex. Water = 4.18 J/(g x ºC)

Glass = 0.50 J/(g x ºC)

Specific Heat• Units for C = J / g ● ºC (joules per gram degree

Celsius)• Equation for Specific Heat: C = q / (m Δ T)• This equation can be rearranged to solve for

heat (q)

q= CmΔT• C = specific heat;• q = heat; • m = mass and • ΔT = change in temperature

Specific Heat

• A 10.0 g sample of iron changes temperature from 25.0C to 50.4 C while releasing 114 joules of heat. Calculate the specific heat of iron.

Example

• C= q/ (m∆T)• C=114 J/ (10.0 g x 25.4°C)• c = 0.449 J/g C

Another example

• If the temperature of 34.4 g of ethanol increases from 25.0 C to 78.8 C how much heat will be absorbed if the specific heat of the ethanol is 2.44 J/g C

Another example

• First, rearrange the specific heat formula to solve for heat

• q = CmT• q = (2.44 J/g°C)(34.4g)(78.8°C –

25.0°C)• q = 4520 J

Yet another example

• 4.50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25.0 C & the specific heat of the gold is 0.129J/g C

Yet another example

• C= q/ (m∆T); rearrange to find• ∆T = q / (C x m)• ∆T = 276 J / (.129 J/g°C x 4.50 g)• T = 475C

• T = Tf-Ti• 475 = Tf - 25• Tf = 500 C