PSC20Heat 1

The Kinetic Molecular Theory(KMT) is used to describe thermal energy,

heat and temperature.KMT is based on 5 postulates.

1) Molecules are in constant random motion, travelling in straight lines

2) Motion is greatest in gases, less in liquids and least in solids.

3) No energy is lost in collisions between molecules

4) molecules in motion have kinetic energy of motion

5) molecules have potential energy due to their position or state

(the farther apart the molecules are, the greater the potential energy)

Thermal Energy – sum of KE and PE of the molecules of the object.

Ethermal = EKE + EPE

Temperature – A measure of the average kinetic energy of the molecules

• not a measure of total KE• is a measure of how fast on average the

molecules are movingHeat (Q) – thermal energy transferred from one object to another due to a temperature difference.

Q = Ethermal• transferred by convection,

conduction or radiation• heat always flows from hot to cold

How Does Heat Move From One Place To Another?

1) Conduction• the movement of heat through solids• hot molecules vibrate faster than cold ones• when hot molecules collide with cooler ones

they transfer some energy causing the cooler ones to vibrate faster

• speed depends on thermal conductivity

2) Convection• transfer of heat by molecules moving

from one place to another• only occurs in liquids and gases

3) Radiation• transfer of energy by electromagnetic

waves• ie. Light, x-rays, microwaves

Which has a higher temperature?Which has more thermal energy?

BLACKSTRAP LAKECANDLE

Blackstrap is definitely colder, but has more thermal energy because there are way more atoms!

Linear Expansion of Solids-solids expand when they are heated and

contract when they are cooled- expansion depends on 3 factors

1) the initial length• the longer the object is to begin with, the more it

will expand2) change in temperature

• the more it is heated, the more it expands3) type of material

• different material will expand different amounts at the same length and temperature change

Linear Expansion

Applications of Linear Expansion

1) Bridges (expansion joints)

2) Railroad Tracks (expansion joints)

This track was laid in February; the picturewas taken in August.

Oil Pipeline Gone Wrong!

L = Li T

L = linear expansion [change in length] (m) = coefficient of linear expansion (oC–1)Li = initial length (m)T = change in temperature (oC)

Sample Problems

1) An unknown metal with a coefficient of linear expansion equal to 38 × 10-6C-1 is 75.1 m long. What is the final length of the metal if the temperature changes from 12.0C to 48.0C?

3) An unknown metal increases in length from 63.200 m to 63.266 m. This happened when the temperature rose from -8.3C to 54.9C. What is the coefficient of linear expansion? What is the metal?

4) What is the initial length of a piece of steel if it is heated from 38.2C to 45.7C and it increases in length by 0.15 cm?

Volume Expansion of Liquids

V = Vi T

V = change in volume (m3) = coefficient of volume expansion (oC –1)Vi = initial volume (m3)T = change in temperature (oC)

Sample Problems1) If the coefficient of volume expansion is 38 ×

10-6 C-1 in a liquid that was originally 45.00 m3

and ended up 45.10 m3, what was the change in temperature? What was the final temperature if the original temperature was 12.0C?

2) What is the coefficient in an unknown liquid if the temperature changes from 6.0C to -25.0C and the liquid shrinks from 7.010 m3 to 6.990 m3?

3) If β = 37 × 10-6 C-1 and the liquid grows 0.10 m3, what is the initial volume when it was heated from -40.0C to 80.0C?

Applications of Volume Expansion

1) Overflow Tanks (Radiators)

2) Thermometers

Measuring Quantities of HeatSpecific Heat Capacity – ability to hold heat• the heat required to raise the temperature of

1 kg of substance by 1oC

Q = m c T

Q = heat (kJ) lost or gainedm = mass (kg)c = specific heat capacity (kJ / kg oC)T = temperature change (oC)

Ex // water = 4.19 kJ / kg oCair = 0.990 kJ / kg oC

Sample Problems

1. How much heat is required to warm 0.100 kg of water from 10.0C to 60.5C?

2. A piece of lead is heated from -12.0C to 9.0C. If it required 3.75 kJ to do this, what is the mass of lead?

3. An unknown metal has a mass of 3850 g. When 14 kJ of heat are applied, the temperature rises from 6.2C to 21.5C. What is the metal?

Principle of Heat Exchange

Heat lost by one object = Heat gained by another- can be used to find specific heat capacity

of an unknown substance-[m1c1ΔT1] = m2c2ΔT2

hot cold

or m1c1ΔT1 + m2c2ΔT2 = 0

Sample Problems1) A 300. g chunk of iron is tossed into a bucket of

methanol. If the original temperature of the iron is 75.00C and the original temperature of the methanol is 4.50C, what would the mass of the methanol present be if the final temperature is 4.80C?

2) A 200. g mass of an unknown metal at 90.0oC is placed in 400. g of water at 20.0oC. The final temperature of the mixture is 23.5oC. What is the specific heat capacity for the metal? What is the metal?

Conservation in Energy Transfer3) A 0.500 kg sample of water is at 15.0C in

a calorimeter. A 0.0400 kg block of zinc at 115.0C is placed in the water. Find the final temperature of the system.

mcTf-Ti + mcTf-Ti = 0(0.500)(4.19)(Tf-15.0) + (0.0400)(0.388)(Tf-115.0) = 0

(2.095)(Tf-15.0) + (0.01552)(Tf-115.0) = 0

2.095Tf – 31.425 + 0.01552Tf – 1.7848 = 0

2.095Tf + 0.01552Tf = 33.2098

2.11052Tf = 33.2098

Tf = 33.20982.11052

Tf = 15.7353… Tf = 15.7C

Another way:Tf = m1C1Ti1 + m2C2Ti2

m1C1 + m2C2

Tf = (0.500)(4.19)(15.0) + (0.0400)(0.388)(115) (0.500)(4.19) + (0.0400)(0.388)

= 31.425 + 1.78482.095 + 0.01552

= 33.20982.11052

= 15.73536…. Tf = 15.7C

4) A 0.250 kg piece of copper at 125.0 C is placed into 0.750 kg of water at 18.0 C. Assuming that no heat is lost to the surroundings, find the final temperature of the system.

Tf = m1C1Ti1 + m2C2Ti2

m1C1 + m2C2

Tf = (0.750)(4.19)(18.0) + (0.250)(0.384)(125.0)(0.750)(4.19) + (0.250)(0.384)

= 56.565 + 12.03.1425 + 0.0960

= 68.5653.2385

= 21.1718…. Tf = 21.2C

Latent Heats

- heat that causes a change of state with no change in temperature

There are two types of latent heats:

Specific Latent Heat of Fusion (lf )

- the quantity of heat needed to melt or freeze 1kg of a solid at its melting/freezing temperature

Specific Latent Heat of Vaporization (lv)

- the quantity of heat needed to vaporize or condense 1kg of a substance at its boiling/condensation temperature

Latent heat is calculated by:

Q = mlf OR Q = mlv

Q = heat , kJm = mass , kgl = specific latent heat, kJ / kg

The molar heat of fusion is similar:

Q = nHfus OR Q = nHvap

Q = heat , kJn = moles, molH = molar heat, kJ / mol

Q = mcΔT

Q = mcΔT

Q = mcΔT Q = mlv

Q = mlf

c = 2.08 kJ/kgºC

c = 2.11 kJ/kgºC

c = 4.19 kJ/kgºC

1) How much heat is needed to vaporize 0.50 kg of water when it has a temperature of 16.5oC?

SAMPLE PROBLEMS

2) How much heat must be lost in order to freeze 150.0 g of water from 45.10C to a temperature of -6.80C?

3) 600.0 kJ of heat is put into a block of ice with a mass of 1.500 kg at a temperature of -15.0C. What is the final temperature of the water?