THE QUANTITY OF HEAT The thermal energy lost or gained by objects is called heat.  One calorie (cal) is the quantity of heat required to change the temperature of one gram of water through one Celsius degree. The mechanical equivalent of heat is given by: 1 cal = 4.186 J

SPECIFIC HEAT CAPACITY The amount of thermal energy required to raise the temperature of a substance varies for different materials.

The specific heat capacity of a material is the quantity of heat required to raise the temperature of a unit mass through one degree.

Q = m c T  Units: Q in J (cal)

m in kg (g), T in K (ºC),c in J/kg.K (cal/g ºC)

Same heat is absorbed.

Iron's ability to store heat is less than water's.

Iron's temperature rises more than does the water's.

Specific Heat Capacity

14.1 How much heat is required to raise the temperature of 200 g of Hg from 20C to 100C?

m = 200 gto = 20C, tf = 100Cc = 0.033 cal/gC

Q = m c T

Q = 200 (0.033) (100 - 20) = 528 cal

MEASUREMENT OF HEAT The principle of thermal equilibrium is the result of a transfer of thermal energy from the warmer bodies to the cooler bodies. If energy is to be conserved we say that the heat lost by the warm bodies must equal the heat gained by the cool bodies.

                              

mw = 80 g, cw = 1 cal/gC, tw= 10Cteq = 18CtCu = 90C, cCu = 0.093 cal/gC

14. 2 A handful of copper shot is heated to 90C and then dropped into 80 g of water at 10C. The final temperature of the mixture is 18C. What was the mass of the shot?

Q LOST(Cu) = Q GAINED (water)

mCu cCu (tCu – teq ) = mw cw (teq – tw )

mm c t t

c t tCuw w eq w

Cu Cu eq

( )

( )mCu

( )( )( )

( . )( )

80 1 18 10

0 093 90 18

= 95.5 g

In this example we have neglected two important facts:o  the water must have a container which will absorb heat from the shoto  the entire system must be insulated from external temperatures otherwise, the equilibrium temperature will always be room temperature

A laboratory device called a calorimeter is used to control these difficulties.

14.3 80 g of dry iron shot is placed in a cup and heated to a temperature of 95C. The mass of the inner aluminum cup and of the aluminum stirrer is 60 g. The calorimeter is partially filled with 150 g of water at 18C. The final temperature of the system is 22C. Find the specific heat capacity of iron.

mFe = 80 g, tFe = 95CmAl = 60 g, cAl = 0.22 cal/gCmw = 150 g, cw = 1 cal/gC, tw= 18Cteq = 22C

Q LOST(Fe) = Q GAINED (water) + Q GAINED (Al)

QGAINED:

Qw = (150)(1)(22-18) = 600 cal

QAl = (60)(0.22)(22-18) = 52.8 cal

Q G(w+Al) = 652.8 cal = Q L(Fe) = mFe cFe T

cQ

m tFeGAINED

Fe

cFe

652 8

80 95 22

.

( )( )= 0.11 cal/gC

CHANGE OF PHASE The change of phase from a solid to a liquid is called fusion, and the temperature at which this change occurs is called the melting point of the substance.The latent heat of fusion Lf of a substance is the heat per unit

mass required to change the substance from the solid to the liquid phase at its melting temperature.

Q = m Lf

The change of phase from a liquid to a vapor is called vaporization, and the temperature at which this change occurs is called the boiling point of the substance.The latent heat of vaporization Lv of a substance is the heat per

unit mass required to change the substance from a liquid to a vapor phase at its boiling temperature.

Q = m Lv

The change of phase from a vapor to a liquid is called condensation.

heat of condensation = heat of vaporization

The change of phase from liquid to solid is called freezing or solidification.

heat of solidification = heat of fusion

The change of phase from solid to gas without passing through the liquid state is called sublimation.

CHANGES OF PHASE

                                                                            

                                                                     

TEMPERATURE vs. QUANTITY OF HEAT

14.4 What quantity of heat is required to change 20 g of ice at -12C to steam at 100C?

mice = 20 g, cice = 0.5 cal/gC, tice = - 12Ccw = 1 cal/gCtsteam = 100CLf = 80 cal/g, Lv = 540 cal/g

Q1: heat needed to raise the tice to its melting point

Q1 = m c T =20(0.5)(0-(-12)) = 120 cal

Q2: heat needed to completely melt the ice

Q2 = m Lf = 20(80) = 1600 cal

mice = 20 g, cice = 0.5 cal/gC, tice = - 12Ccw = 1 cal/gCtsteam = 100CLf = 80 cal/g, Lv = 540 cal/gQ1: heat needed to raise the tice to its melting point

Q1 = m c T =20(0.5)(0-(-12)) = 120 cal Q2: heat needed to completely melt the ice

Q2 = m Lf = 20(80) = 1600 cal

Q3: heat needed to raise the tw to its boiling point

Q3 = m c T =20(1)(100-0) = 2000 cal

Q4: heat needed to completely vaporize the water

Q4 = m Lv = 20(540) = 10,800 cal

QTOTAL = 120 + 1600 + 2000 + 10800

= 14,520 cal

14.5 After 12 g of crushed ice at -10C is dropped into a 50 g aluminum calorimeter cup containing 100 g of water at 50C, the system is sealed and allowed to reach thermal equilibrium. What is the resulting temperature?

mice = 120 g , cice = 0.5 cal/gCmAl = 50 g, cAl = 0.22 cal/gCmw = 100 g, cw = 1 cal/gCtice = - 10C, tw = 50CLf = 80 cal/g

Q LOST = Q GAINED

Q LOST = Q (water) + Q (Al)

Q GAINED = Q (ice) + Q (fusion) + Q (water)

mice = 120 g , cice = 0.5 cal/gCmAl = 50 g, cAl = 0.22 cal/gCmw = 100 g, cw = 1 cal/gCtice = - 10C, tw = 50CLf = 80 cal/g

Q LOST = mw cw T + mAl cAl T

100(1)(50- teq ) + 50(0.22)(50- teq) = 5550 - 111 teq

Q GAINED = mice cice T + mice Lf + mw cw T

12(0.5)(0 - (-10)) + 12(80) + 12(1) (teq - 0) = 1020 + 12 teq

5550 - 111 teq = 1020 + 12 teq

teq = 36.8C

14.6 If 10 g of steam at 100C is introduced into a mixture of 200 g of water and 120 g of ice, find the final temperature and composition of the mixture.

mice = 120 g, cice = 0.5 cal/gCmw = 200 g, cw = 1 cal/gCmsteaml = 10 g, tsteam = 100CLf = 80 cal/g, Lv = 540 cal/g

Heat needed to melt the ice: Q1 = mice Lf

= 120(80) = 9600 cal

Heat given off by steam : Q2 + Q3 = msteam Lv + mw cw T

= 10(540) + 10(1)(100 - 0) = 6400 cal

mice = 120 g, cice = 0.5 cal/gCmw = 200 g, cw = 1 cal/gCmsteaml = 10 g, tsteam = 100CLf = 80 cal/g, Lv = 540 cal/g

Heat available = 9600 - 6400 = 3200 cal

mice Lf = 3200 cal

mice = 3200/80 = 40 g

mw = initial mass + melted ice + condensed steam

 mw = 200 + (120 - 40) + 10 = 290 g

Final temperature must be 0C