Unit 12

Thermochemistry

THERMOCHEMISTRY

A branch of thermodynamics which focuses on the

study of heat given off or absorbed in a chemical

reaction.

Heat absorbed: A + B + heat C

Heat given off: A + B C + heat

Reactants

Reactants

Products

Products

HEAT is the transfer of energy from one body to another as a

result of a difference in temperature or a change in

phase.

• Expressed as Q in math equations.

• Units used to measure heat is joules or kilojoules

Direction of Heat or Q.

Excited atoms’ vibration (more

energy) cause atoms nearby to

also vibrate vigorously.

Therefore, you can say heat

transfer from hot to cold. Temperature:

100 C

Temperature 0 C

Heat from the body is warmer than

the temperature of the surrounding.

Therefore, heat will transfer

towards the direction that is colder.

The person may lose enough body

temperature (heat) and can

experience hypothermia.

EXAMPLES

Heat from the body is transferred to

the ice pack so that the injured

muscle tissues can heal.

Heat

EXAMPLES The temperature that is surrounding

the icy water is warmer than the ice

water. Therefore, the heat will

transfer from the surrounding

(warmer) to the ice water (colder).

One observes the ice (solid water)

changing phase (liquid water) or

melting because heat has moved in.

And also one observes

condensation which is water vapor

from the surrounding air (gaseous

water) changing phase (liquid

water) on the glass because the

glass temperature is colder than the

surrounding (warmer).

TEMPERATURE is the measure of the average kinetic energy (KE) of the

particles in a sample of matter.

• Expressed as T in math equations

• or ΔT – change in temperature.

• Units used Celsius (ºC) or Kelvin (K).

Lower temperature means the particles have a lower average

kinetic energy. Higher temperature means the particles have

higher average kinetic energy.

inc

temp

inc

KE

dec

temp

dec

KE

HEAT vs TEMPERATURE

Heat is a measure

of transferred

energy.

Temperature is a

measure of average

kinetic energy

contained in that

matter.

The End of Part 1

SPECIFIC HEAT is the quantity of heat required to raise the temperature

of a substance by one degree Celsius

• Expressed as c in math equations

897 j/kg·K

(specific heat)

is required to

raise the

temperature of

aluminum by

1ºC.

Different materials heat up at different rates

(different specific heat).

Example of different SPECIFIC HEAT

So which material has the highest specific heat?

So which material has the highest specific heat?

Water has the highest specific heat on this chart.

That is why water is used sometimes as coolant because it

takes a lot of heat (4.18 j/g·ºC) in order for it to only raise

its temperature by 1ºC.

• Water and lead do not transfer heat at the same value.

Water has a specific heat (C) = 4.18 J/goC

lead has a specific heat (C) = 0.13 J/goC

• What does their specific heat numbers mean?

It means it requires 4.18 Joules of energy to heat 1 gram of

water to 1oC more and only 0.13 Joules of energy to heat 1

gram of lead to 1oC more.

Water heats up slowly and requires a lot of energy whereas

lead does not.

In other words…

if you compare water to lead,

That is why water is used sometimes as coolant because it

takes a lot of heat (4.18 j/g·ºC) in order for it to only raise

its temperature by 1ºC.

Calculations involving SPECIFIC HEAT Formulas:

Q = m c ΔT or c = Q

m ΔT Heat

transferred

mass specific

heat Change

in temp

Q = energy transferred (joules)

m = mass (grams)

c = specific heat (j/g·ºC or j/g·K)

Δ T = temperature change (K or ºC)

Tfinal – T initial

Example 1- SPECIFIC HEAT problems:

How much heat is required to be absorbed to raise

the temperature of 6 g of water, 5 ˚C?

(specific heat of water is 4.18 J/ g·˚C)

The question looks complicated at first so break the process

down to solve.

Example SPECIFIC HEAT problems:

How much heat is required to be absorbed to raise

the temperature of 6 g of water, 5 ˚C?

(specific heat of water is 4.18 J/ g·˚C)

First, circle what is it that they are asking for and then underline the

given data.

Example SPECIFIC HEAT problems:

How much heat is required to be absorbed to raise

the temperature of 6 g of water, 5 ˚C?

(specific heat of water is 4.18 J/ g·˚C)

Second, decide which formula to use by looking at the what you have

circled or what they are asking for.

Q = m c ∆T or c = Q

m ∆T This stands for heat. This stands for

specific heat.

Example SPECIFIC HEAT problems:

How much heat is required to be absorbed to raise

the temperature of 6 g of water, 5 ˚C? (specific

heat of water is 4.18 J/ g·˚C)

They are asking about heat (what we circled) so we will use the first

formula.

Q = m c ∆T or c = Q

m ∆T This stands for heat. This stands for

specific heat.

Example SPECIFIC HEAT problems:

Q = m c ∆T

Q = (6g) (4.18 j/g●°C) (5 °C)

Q = 125.4 j

How much heat is required to be absorbed to raise

the temperature of 6 g of water, 5 ˚C?

(specific heat of water is 4.18 J/ g·˚C)

So in the calculator, it should read 6 · 4.18 · 5 =

Δ T = Tfinal – Tinitial

Δ T = 5 ºC – 0 ºC

= 5 ºC

Example 2- SPECIFIC HEAT problems:

Find the specific heat of a material if a 5 gram sample

absorbs 50 J when it is heated from 30˚C to 50ºC.

Remember to circle the unit that they are asking for and underline the given

data.

Example 2- SPECIFIC HEAT problems:

Find the specific heat of a material if a 5 gram sample

absorbs 50 J when it is heated from 30˚C to 50ºC.

Next, decide on the formula to use which depends on what it was asked.

Q = m c ∆T c = Q

m ∆T or

This stands for

specific heat.

Example 2- SPECIFIC HEAT problems:

Find the specific heat of a material if a 5 gram sample

absorbs 50 J when it is heated from 30˚C to 50ºC.

On the calculator, it should look like 50/(5 · 20) =

c = Q

m ∆T

C = 50 j

5 g • 20 ºC

Δ T = 50 ºC – 30 ºC

= 20 ºC

C = 0.50 j/g·ºC

The End of Part 2