THE QUANTITY OF HEAT The thermal energy lost or gained by objects is called heat.

1. 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

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

3. 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

4. 14.1 How much heat is required to raise the temperature of 200 g of Hg from 20C to 100C? m = 200 g to = 20C, tf = 100C c = 0.033 cal/gC Q = m c T Q = 200 (0.033) (100 - 20) = 528 cal

5. 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.

6. 14. 2 A handful of copper shot is heated to 90C and then dropped into 80 g of water at 10C. The final temperature of the mixture is 18C. What was the mass of the shot? mw = 80 g, cw = 1 cal/gC, tw= 10C teq = 18C tCu = 90C, cCu = 0.093 cal/gC Q LOST(Cu) = Q GAINED (water) mCu cCu (tCu – teq ) = mw cw (teq – tw ) = 95.5 g

7. In this example we have neglected two important facts: othe water must have a container which will absorb heat from the shot othe 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.

8. 14.3 80 g of dry iron shot is placed in a cup and heated to a temperature of 95C. 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 18C. The final temperature of the system is 22C. Find the specific heat capacity of iron. mFe = 80 g, tFe= 95C mAl = 60 g, cAl = 0.22 cal/gC mw = 150 g, cw = 1 cal/gC, tw= 18C teq = 22C 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

9. Q G(w+Al) = 652.8 cal = Q L(Fe) = mFe cFe T = 0.11 cal/gC

10. 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 Lfof 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

11. The change of phase from a liquid to a vapor is calledvaporization, and the temperature at which this change occursis called the boiling point of the substance. Thelatent heat of vaporization Lvof 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

12. The change of phase from a vapor to a liquid is calledcondensation. heat of condensation = heat of vaporization The change of phase from liquid to solid is calledfreezingor solidification. heat of solidification = heat of fusion The change of phase from solid to gas without passing through the liquid state is calledsublimation.

13. CHANGES OF PHASE

14. TEMPERATURE vs. QUANTITY OF HEAT

15. 14.4 What quantity of heat is required to change 20 g of ice at -12C to steam at 100C? mice = 20 g, cice = 0.5 cal/gC, tice = - 12C cw = 1 cal/gC tsteam = 100C Lf = 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

16. mice = 20 g, cice = 0.5 cal/gC, tice = - 12C cw = 1 cal/gC tsteam = 100C Lf = 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 Q3: heat needed to raise the twto 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

17. 14.5 After 12 g of crushed ice at -10C is dropped into a 50 g aluminum calorimeter cup containing 100 g of water at 50C, the system is sealed and allowed to reach thermal equilibrium. What is the resulting temperature? mice = 120 g , cice = 0.5 cal/gC mAl = 50 g, cAl = 0.22 cal/gC mw = 100 g, cw = 1 cal/gC tice = - 10C, tw = 50C Lf = 80 cal/g Q LOST = Q GAINED Q LOST = Q (water) + Q (Al) Q GAINED = Q (ice) + Q (fusion) + Q (water)

18. mice = 120 g , cice = 0.5 cal/gC mAl = 50 g, cAl = 0.22 cal/gC mw = 100 g, cw = 1 cal/gC tice = - 10C, tw = 50C Lf = 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 = 36.8C

19. 14.6 If 10 g of steam at 100C 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/gC mw = 200 g, cw = 1 cal/gC msteaml = 10 g, tsteam = 100C Lf = 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

20. mice = 120 g, cice = 0.5 cal/gC mw = 200 g, cw = 1 cal/gC msteaml = 10 g, tsteam = 100C Lf = 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 0C