heat and temperature heat is a form of energy, and is measured in joules (j). temperature is...

Heat and Temperature

Heat is a form of energy, and is measured in Joules (J).

Temperature is different from heat.

Temperature is a measure of how hot or cold an object is.

It is measured in degrees Celsius (°C).

Heat can move from one place to another.

The three methods of heat travel are:

• conduction

• convection

• radiation

barrier

HEAT

Conduction

Experiment

A B C

Pins A, B and C are held in place by vaseline/wax.

The barrier stops heat reaching the pins through the air.

CONDUCTION occurs in solids when heat passes from particle to particle along the solid.

Convection

Experiment

water

purple crystal

The purple colour allows us to see the heated water moving.

HEAT

The convection currents look like this:

hot water rising

cold water falling

CONVECTION occurs in liquids and gases because their particles can move.

Heated air and liquid will rise upwards.

Day Break

Land is warmer than sea.

Air heats up and rises over land, cools down and falls

over sea.

Sunset

Sea is warmer than land.

Air cools and falls over land, but is warmed over the sea

and rises.

Radiation

Experiment

A metallic cube known as a “Leslie Cube” is filled with boiling water.

All hot objects radiate heat energy.

Radiated heat waves are also known as INFRARED RADIATION.

It travels at the speed of light (3 x 108 ms-1).

Matt black surfaces radiate more heat than shiny surfaces at the same temperature.

Infrared radiation can travel through a vacuum (just as heat from sun is able to travel through space which is a vacuum).

Nocturnal Lemur

Black fur poses no disadvantage as only

active at night

Lemur

Active during the day so has lighter fur.

On cold mornings points its belly to the sun.

Heat Loss

Reducing Heat Loss

Preventative Action Heat Lost Type

cavity wall insulation conduction

double glazing conduction

carpets conduction

loft insulation convection

draft excluders convection

shiny foil behind radiators radiation

Heat energy is lost faster when there is a difference in temperature between inside and outside.

bigger

Heat Energy

The amount of heat energy required to heat up an object depends on:

• the mass of the substance ( m )

• the material the object is made of

• the rise in temperature required ( ΔT )

The amount of heat energy is given by: ΔT m cEH

x

÷EH

c m ΔT

Quantity Unit

Heat Energy ( EH )

Specific Heat Capacity ( c )

Mass ( m )

Change in Temperature ( ΔT )

Joules ( J )

Joules per kilogram per degrees Celsius ( J kg-1 °C-

1 )

Kilograms ( kg )

Degrees Celsius ( °C )

** Specific Heat Capacities can be found on Data Sheet in exam paper **

Specific Heat Capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C.

Example 1

Calculate the amount of heat energy required to raise the temperature of 2 kg of water from 20 °C to 100 °C.

kg 2m

?EH

20100ΔT C 80

-1-1 C kg J 4180c

ΔT m cEH

80 2 4180

J 668,800EH

** data sheet **

Example 2

A 3 kg block of copper at 15 °C is heated.

Calculate the rise in temperature if 20 kJ of heat energy is transferred to the copper.

kg 3m

kJ 20EH J 10 20 3 J 20,000

-1-1 C kg J 380c

?ΔT

m cE

ΔT H

33801020 3

C 17.5ΔT

17.515etemperatur final

C 32.5

** data sheet **

Example 3

0.5 kg of a liquid is heated from 8 °C to 40 °C by 50 kJ of heat energy.

Calculate the specific heat capacity of the liquid.

kg 0.5m

kJ 50EH

840ΔT C 32

?c

J10 50 3 J 50,000

ΔT mE

c H

320.51050 3

-1-1 C kg J 3,125c

Yellow Book

Heat Energy – Page 90

Q79 (a) (b) (c) and (d)

Q80, Q81, Q82, Q83, Q84, Q85

Power of a Heater

Experiment

Joulemeter

12 volt Immersion

Heaterto 12 V

a.c. supply

stopwatch

Zero the joulemeter.

Switch on for 20 seconds and record the energy used.

s 20t

J E

?P

tE

P

20

W P

Specific Heat Capacity

Experiment

The specific heat capacity of aluminium is measured using the apparatus shown.

thermometer

aluminium block

heater

to 12 V a.c.

supply

stopwatch

Initial temperature of block °C.

The heater is switched on for 5 minutes.

Final temperature of block °C.

min 5t

Heat Energy Used

s 300

W P

?EH

tPEH

300

J EH

Specific Heat Capacity

C etemperatur initial

C etemperatur final

C ΔT

J EH

kg m

?c

ΔT mE

c H

-1-1 C kg J c

The value measured experimentally is higher than 380 J kg-1 °C-1.

This is because not all the heat energy is transferred to the aluminium block – some is lost to the surroundings.

Since E = P x t we can write P x t instead of E.

Combining EH = cmΔT and E = Pxt

The heat energy equation is:

The equation for electrical power is:

ΔT m cEH

tPE

ΔT m cEH

ΔT m ctP ** NOT on data sheet **

-1-1 C kg J 4180c

Example 1

2 kg of water is heated from 20°C to 60°C by a 2 kW heater.

Calculate the time taken.

kg 2m

kW 2P

2060ΔT C 40

W 2,000

?t

ΔT m cEH

ΔT m ctP

40 2 4180t 2,000

200040 2 4180

t

2,000334,400

t

s 167.2t

Questions

Q1. 0.5 kg of copper is heated from 15 °C to 50 °C by a 60 W heater.

Calculate the time taken.

Q2. 2 kg of water at 10 °C is heated by a 1.5 kW heater for 3 mins.

Calculate the final temperature of the water.

Q3. How long will it take a 2.3 kW heater to boil 0.5 kg of water with an initial temperature of 10 °C.

112.6 s

ΔT = 32.3 °C so Tfin = 42.3 °C

81.8 s

Measuring & Calculating Time Taken

Experiment

The time taken for a kettle to boil is measured and calculated.

W 1,850P

C Tinitial

C 100Tfinal C ΔT

-1-1water C kg J 4,180C

ΔT m ctP

1 4,180t1,850 kg 1m

t1,850

1,850

t

s t

The measured time taken for the kettle to boil is s.

Variations of EH = cmΔT

There are several formulae for energy.

These can be used to do heat calculations.

ΔT m cEH

ΔT m ctP

ΔT m ctIV

ΔT m cv m 21 2

** NOT on data sheet **

Example 1

A heater operates from the 230 V mains supply and takes a current of 3 A.

This heater is used to heat up 2 kg of water at 18 °C.

The heater is switched on for 5 minutes.

Calculate the rise in temperature of the water and then state the final temperature reached.

V 230V

A 3I kg 2mmins 5t

560s 300

?ΔT

-1-1water C kg J 4,180C

ΔT m ctIV

ΔT 24,1803003230

ΔT 8,360207,000

207,000ΔT 8,360

8,360207,000

ΔT

C 24.8ΔT

The final temperature of the water is: 24.818Tfinal C 42.8

Example 2

A lead bullet of mass 0.05 kg has a speed of 80 ms-1.

The bullet hits a wall and stops.

Calculate the rise in temperature of the bullet as it does this.

(clead = 130 J kg-1 °C-1)

All the kinetic energy is transferred into heat energy (EK = EH)

kg 0.05m-1ms 80v

-1-1lead C kg J 130c

?ΔT

ΔT m cv m 21 2

ΔT 0.05130800.05 21 2

ΔT 6.5160

160ΔT 6.5

6.5160

ΔT

C 24.6ΔT

Question

A 2.2 kW kettle contains 1.7 kg of water at 12 °C.

The kettle is switched on for 4 minutes.

(a) Calculate the maximum temperature reached by the water.

(b) In practice, will the maximum temperature reached be bigger or smaller than this calculated value.

Explain your answer fully.

kW 2.2PW 2,200

mins 4t 460s 240

kg1.7 m-1-1

water C kg J 4,180c

?ΔT

ΔT m ctP

ΔT 1.741802402,200

ΔT 7,106528,000

528,000ΔT 7,106

7,106528,000

ΔT

C 74.3ΔT

The maximum temperature of the water is: 74.312Tmax C 86.3

(a)

(b) In practice, the maximum temperature reached will be less than 74.3 °C.

This is because some heat energy is lost to the surroundings.

Cooling Salol

Experiment

Liquid salol at 60°C is allowed to cool.

The temperature of the liquid salol is measured every minute as it cools.

liquid Salol

thermometerstopwatch

Results

Time(minutes)

Temperature(°C)

123...

25

A graph of the temperature against time is plotted for the cooling salol.

A

B C

D E

temperature

(°C)

time

38

change of state

During a change of state – NO CHANGE IN TEMPERATURE.

A

B C

D E

temperature

(°C)

time

38

Stage Description

AB liquid cools down

BC

CD

liquid salol turns to a solid (solidifies)latent (hidden) heat is given out

temperature stays the same

solid salol cools down to room temp.

salol is at room temp. so no further fall in temperature

DE

Change of State

The changes of state are:

Solid Liquid Gas

melting evaporation

condensationfreezing

Latent Heat of Fusion (melting)

The amount of heat energy required to melt a solid depends on:

1. the mass of the solid

2. what substance the solid is made of.

This gives the equation:fH L m E

Quantity Unit

Heat Energy ( EH )

Mass ( m )

Specific Latent Heat of Fusion ( Lf )

Joules ( J )

Kilograms ( kg )

Joules per Kilogram ( J kg-1 )

x

÷EH

m Lf

The specific latent heat of fusion for water is 3.34 x 105 J kg-1.

Specific Latent Heat of Fusion is the amount of energy required to change 1 kg of a substance from solid to liquid

Example 1

Calculate the energy required to melt 2 kg of ice at 0 °C.

?EH

kg 2m-15

f kg J 103.34L fH L mE

5103.34 2

J106.68E 5H

Example 2

Calculate the time taken by a 500 W heater to completely melt 2 kg of ice at 0 °C.

kg 2m

W 500P

-15f kg J 103.34L

fH L mE

fL mtP

5103.34 2t 500

500106.68

t5

s 1,336t

Questions

1. Calculate the heat energy required to melt 2.7 kg of ice at 0 °C.

2. A 700 W heater inside a block of ice at 0 °C is switched on for 3 minutes.

Calculate the mass of ice which will melt.

3. A 1 kW heater inside a block of ice at 0 °C is switched on.

The heater melts 0.65 kg of ice.

Calculate the time that the heater was switched on for.

Measuring Specific Latent Heat of Fusion of Ice

Experiment

Joulemeter

Ice at 0 °CHeater

(not switched on)

Beaker 2

Control Experiment

Water

Beaker 1

The control experiment is not switched on.

The control measures the amount of ice that melts due to temperature of the room.

The heater connected to the joulemeter is switched on for 5-minutes.

The amount of ice melted in beaker 2 is subtracted from the amount melted in beaker 1.

Results

J EH

kg m room heater by melted

kg m room by melted

kg m only heater by melted

fH L mE

mE

L Hf

-1f kg J L

The data booklet value is 3.34 x 105 J kg-1.

Q. Why is the measured value less than the data book vale?

The measured value is less than 3.34 x 105 J kg-1 as heat energy is lost to the surroundings.

Latent Heat of Vaporisation

The heat energy needed to boil away a liquid (vaporise) depends on:

1. the mass of liquid

2. what the liquid is.

This gives the equation:

The specific latent heat of vaporisation of water is 2.26 x 106 J kg-1.

VH L m E

Questions

1. Calculate the energy required to turn 3kg of water at 100 °C into steam at 100 °C.

2. Calculate the time taken by a 2 kW kettle to boil away 0.5 kg of water at 100 °C.

Measuring Specific Latent Heat of Vaporisation

Experiment

water

1,850 W heating element

The mass of the kettle and water is measured.

The kettle is kept on for 5 minutes while the water boils.

The new mass of the kettle and boiling water is taken.

Results

kg m water kettle

kg m water boiling kettle

kg m steam to turned water

W 1,850 P

mins 5 tboiling 60 5

s 300

VH L mE

VL mtP

VL 3001,850

555,000L V

555,000LV

-1V kg J L

Applications Involving a Change of State

Cool Box

frozen pack

• A frozen pack is placed inside the cool box.

• It absorbs heat energy from the air and the food inside.

• The frozen pack melts ( solid to liquid ).

• The air and food are cooler.

M

compressor

Heat

The Fridge

Heat

• the motor pumps a liquid through pipe work inside fridge.

• the liquid absorbs (latent) heat energy, and changes to a gas.

• air inside fridge becomes cooler.

• compressor changes gas back to a liquid.

• gas loses heat energy as travels through pipes at rear.

Complex Heat Calculations

Example 1

(a) Calculate the heat energy required to turn 0.5 kg of ice at 0°C into water at 20°C.

cwater = 4180 J kg-1 °C-1 Lf ice = 3.34 x 105 J kg-1

There are 2 parts to this problem:

1. melting ice

2. heating water

melting the ice

?EH

kg 0.5m-15

f kg J 103.34L fH L mE

5103.34 0.5

J101.67E 5H (167,000 J)

heating the water

kg 0.5m

?EH

020ΔT C 20

-1-1 C kg J 4180c

ΔT m cEH

20 0.5 4180

J 41,800EH

41,800 101.67 energy total 5 J 10 2.088 5

(b) How long would it take a 700 W kettle to do this?

tE

P

x

÷E

P t

W 700P

J 102.088E 5

?t

PE

t

700102.088 5

s 298t

Questions

1. (a) Calculate the heat energy required to turn 1.2 kg of ice at 0°C into water at 18°C.

(b) Calculate the time it would take a 600 W heater to do this.

400,800 + 90,288 = 491,088 J

818.5 seconds

time (s)

temperature (°C)

10

40

50 250

2. A 0.2 kg mass of a solid substance is heated by a 150 W heater.

A graph of temperature against time for the substance as it is heated is shown.

(a) What happens between 50 and 250 seconds?

(b) What is the melting point (temperature) of the solid?

(c) Calculate the specific heat capacity of the solid.

(d) Calculate the specific latent heat of fusion of the substance.

change of state

40 °C

1250 J kg-1 °C-1

150,000 J kg-1`