II. Heat

1. Heat and energy

Calorie: the amount of heat required to raise the temperature

of 1 gram of water from 14.5 ◦C to 15.5 ◦C.

A little bit of history:• Old theories• Robert von Mayer (1814-1878)• James Joule (1818-1889)• Hermann von Helmholtz (1821-1894)

1 cal = 4.186 J

2. Specific heat

dT

dQ

mc

1

KkgJCgcalcwater /4190/1

1 cal = 4.186 J

cmdTdQ TcmQ

2a. Heat capacity

dT

dQC CdTdQ TCQ cmC

2. CalorimetryΔQ=cmΔT

0iQ

Example: 200 g of water at the temperature of 80°C is placed into a 50 g glass at 20°C. What is the final temperature of the system.

mw = 200g Tw = 80º Ccw=1.0cal/(gCº)mg = 50g Tg = 20º Ccg = 0.20cal/(gCº)

T=Tf - ?

0 gw QQ

0)( wwww TTmcQ

0)( gggg TTmcQ

0)()( gggwww TTmcTTmc

ggww

gggwww

mcmc

TmcTmcT

0iQ 0)( iii TTmc

ii

iii

mc

TmcT

3. Phase transitions.

Q = ±mL ΔQ = ±ΔmL

Example: 100 g of ice at the temperature of -40°C should be melted and heated to the temperature of 20°C. Calculate needed heat.

calgcalg

gcal

gQ

12000/]208020[*100

)]Cº0Cº20(*)º1.0cal/(gC

/80

))Cº40(Cº0(*)º0.5cal/(gC[*100

m= 100g Ti = -40º Cci=0.5cal/(gCº)T = 20º Ccw =1.0cal/(gCº)Tc = 0º CL = 80 cal/g

ΔQ - ?

)()(c

)()(mc

cwici

cwici

TTcLTTm

TTmcmLTTQ

Latent heat: energy required for a material to change phase, even though its temperature is not changing.

Example: Ice and Water

mw = 300g mi = 200gTw = 20º CTi = -40º Cci=0.5cal/(gCº)cw =1.0cal/(gCº)Tc = 0º CL = 80 cal/g

C0T

25gm

0cal/g 8040006000

cal/g 80m

4000400200/5.0)(mc

6000200300/0.1)(

i

mcalcalQQQ

mLQ

calCCggCcalTTQ

calCCggCcalTTmcQ

iwiw

iw

icii

wcwww

200 g of ice at -40º C is mixed with 300 g of water at 20º C.What will happen with ice and water?

25 g of ice will be meltedThe temperature of the mixture will be 0º C

Energy is required for a material to change phase, even though its temperature is not changing.

Example

Molar specific heat (molar heat capacity)

dT

dQ

ncM

1ndTcdQ MTncQ M

One mole is the amount of substance that contains as many elementary entities as there are atoms in 12g of carbon -12

molmoleculesN A /10022.6 23

nmnNmNm Asample

AmN

AN

Nn

Molecular weight

n - number of molesN - number of moleculesNA - Avogadro number msample - total mass of samplem - mass of one moleculeμ=M - molar (atomic) mass (“weight”)

Avogadro number and number of moles

sampleM cmnc ccM

A beaker contains 0.200 kg of water. The heat capacity of the water is ___ J/˚C.

Assume: the specific heat of water is 4000 J/(kg·˚C).

Question

1. 200

2. 400

3. 600

4. 800

The heat capacity of the water in a massless beaker is 800 J/˚C. The amount of heat needed to increase the temperature of the water from 15.0 ˚C to 30.0 ˚C is ___ kJ.

Question

1. 42. 8

3. 12

4. 16

A liter of water is put in a sauce pan on a stove. In order to heat the water at 10 ˚C/minute requires a heating power of ___ Watts.

(Ignore the heat capacity of the pan. Assume the heat capacity of a liter of water is 4200 J/˚C.)

Question

1. 500

2. 700

3. 900

4. 1100

After the blocks are put into thermal contact and reach equilibrium, the final temperature is __ ˚C.

Question

1. 50

2. 30

3. 70

4. 40

Block A: 1.0 kg, c = 400 J/(kg·˚C), initial temperature = 100 ˚C

Block B: 2.0 kg, c = 200 J/(kg·˚C), initial temperature = 0 ˚C

Example

How much heat does it take to heat 1.00 kg of water ice from 10 ˚C up to 100 ˚C and boil the water away as steam?

Given: cice 2.10 kJ/kg·°C

cwater 4.19 kJ/kg·°C

Lf 334 kJ/kg Lv 2256 kJ/kg

1. Heat the ice up to 0 °C

Q1 mciceT (1 kg)(2.1 kJ/kg·°C)(10 °C)

= 21.0 kJ.

2. Melt the ice at 0 °C

Q2 mLf (1 kg)(334 kJ/kg) = 334 kJ

3. Heat the liquid water up to 100 °C

Q3 mwatercwaterT

(1 kg)(4.19 kJ/kg·°C)(100 °C)

= 419 kJ

4. Boil off the water into steam at 100 °C

Q4 mLv (1 kg)(2256 kJ/kg) = 2256 kJ

Total

Q Q1 Q2 Q3 Q4

= 21 kJ + 334 kJ + 419 kJ + 2256 kJ

= 774 kJ + 2256 kJ

Q = 3030 kJ = 3.03 MJ

Most (~ 3/4) of the heat is used in boiling

off the water.

ExampleA 10 g ice cube at 0 ˚C is put into 100 g of water at 20 ˚C. Does the ice cube melt completely? If not, the final temperature of the water + ice cube would be 0 ˚C. If so, what is the final temperature of the water?

Lf 334 kJ/kg cwater 4.19 kJ/kg·°C

The heat needed to be added to the ice to

melt it is

Qice miceLf (0.010 kg)(334 kJ/kg) = 3.34 kJ.

The heat given off by the liquid water when it

is cooled to 0 °C is

| Qwater | mwaterc | T |

(0.100 kg)(4.19 kJ/kg·°C)(20 °C)

= 8.38 kJ.

Thus there is more than enough thermal

energy in the water to completely melt the ice.

The heat absorbed by the ice to raise its

temperature to the final temperature T2 is

Qice miceLf micecwater (T2 Tice 1)

miceLf micecwaterT2 .

The heat absorbed by the water to decrease

its temperature to the final temperature is

Qwater mwatercwater (T2 Twater 1).

The sum of these two heats adds to zero:

miceLf micecwaterT2 mwatercwater (T2 Twater 1) 0.

Solving for T2 gives

T2 mwatercwaterTwater 1 miceLf

(mice mwater )cwater

= (0.100 kg)(4.19 kJ/kg·°C)(20 °C) 3.34 kJ

(0.010 kg + 0.100 kg)(4.19 kJ/kg·°C)

= 10.9 °C .