2. WHAT ARE SOME OF THE WAYS THAT HEAT ENERGY CAN BE TRANSFERRED? 1) CONDUCTION – IF TWO OBJECTS ARE IN CONTACT, THERMAL ENERGY CAN BE TRANSFERRED THROUGH COLLISIONS OF THE ATOMS/MOLECULES OF THE OBJECTS AT THE INTERFACE BETWEEN THE TWO OBJECTS. CONDUCTION DOES NOT INVOLVE MATERIAL TRANSFER.

3. 2) CONVECTION IS THE PROCESS OF HEAT TRANSFER FROM ONE LOCATION TO ANOTHER BY THE MOVEMENT OF FLUIDS. AS A FLUID IS HEATED, IT BECOMES LESS DENSE AND RISES. COOLER FLUID WOULD TAKE ITS PLACE.

4. 3) RADIATION IS THE TRANSFER OF HEAT THROUGH ELECTROMAGNETIC RADIATION. AS THE TEMPERATURE OF AN OBJECT INCREASES, THE AMOUNT OF RADIATION ALSO INCREASES. RATE OF RADIATION = k x T 4 THE AMOUNT OF RADIATION INCREASES WITH INCREASE IN TEMPERATURE. THE WAVELENGTHS OF EMISSION ALSO CHANGE WITH THE WAVELENGTH OF MAXIMUM EMISSION SHIFTING TO SHORTER WAVELENGTH AS THE TEMPERATURE INCREASES.

6. SUPPOSE THAT OBJECT A AND OBJECT B HAVE REACHED A THERMAL EQUILIBRIUM. DO THE PARTICLES OF THE TWO OBJECTS STILL COLLIDE WITH EACH OTHER? IF SO, DO ANY OF THE COLLISIONS RESULT IN THE TRANSFER OF ENERGY BETWEEN THE TWO OBJECTS? EXPLAIN.

7. OBJECTS DO NOT CONTAIN HEAT. OBJECTS ARE MADE UP OF ATOMS AND MOLECULES AND THEY CONTAIN ENERGY (THERMAL ENERGY). HEAT IS THE TRANSFER OF ENERGY FROM AN OBJECT TO ANOTHER OBJECT OR TO ITS SURROUNDINGS. SO, WHAT DOES HEAT DO?

8. 1)HEAT CHANGES THE TEMPERATURE OF AN OBJECT. 2)HEAT CHANGES THE STATE OF AN OBJECT. 3)HEAT CAN CAUSE WORK TO BE DONE.

9. UNITS FOR ENERGY ARE JOULES, ALTHOUGH CALORIES ARE SOMETIMES USED. 1 CALORIE = 4.184 JOULES IF WE HEAT DIFFERENT SUBSTANCES AT THE SAME RATE, THE TEMPERATURE CHANGES WE WILL SEE WILL BE DIFFEREN. DIFFERENT SUBSTANCES HAVE DIFFERENT HEAT CAPACITIES. HEAT CAPACITY – THE HEAT REQUIRED TO RAISE THE TEMPERATURE OF 1 GRAM OF A SUBSTANCE BY 1 DEGREE CELCIUS.

10. THE EQUATION RELATING THE AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF A GIVEN AMOUNT OF MATERIAL IS GIVEN AS: Q = m x C x  T Where m = mass C = heat capacity  T = change in temperature Q = heat transferred

11. What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C. Q = m x C x  T m = 450 g C = 4.18 J/g/ o C  T = 85 – 15 = 70 o q Q = 450g x 4.18 J/g/ o C x 70 o = 131670 J = 131.7 kJ

12. A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.

13. IF THEY COME TO THERMAL EQUILIBRIUM, THE HEAT LOST BY THE WATER = THE HEAT GAINED BY THE METAL Q water = 50 g x 4.18 x (88.6 – 87.1) = 50 x 4.18 x 1.5 = 313.5 J Q metal = 12.9 x C m x (87.1 – 26.5) 313.5 = 12.9 x C m x 60.6 Or C m = 313.5/782 = 0.402 J/g/ o C

14. FOR CHANGES IN STATE, YOU HAVE SIMILAR SITUATIONS. TO MELT SOMETHING, YOU HAVE TO PUT ENERGY IN. TO FREEZE SOMETHING, ENERGY HAS TO BE LOST OR REMOVED. Q fusion = m x  H fusion FOR VAPORIZATION (BOILING) AND CONDENSATION Q vap = m x  H vap

15. Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g.

16. Q fusion = m x  H fusion = 48.2 g x333 J/g = 16051 J = 16.1 kJ

17. What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g.

18. FIRST, LET’S MELT THE ICE. Q fusion = 50 x 333 J/g = 16650 J NOW, THE WATER CAN COOL FROM 26.5 o TO 0 o Q water = m x 4.18 x 26.5 So 16650 = m x 111 Or m = 16650/111 = 150 g

19. IF YOU ARE GOING TO MAKE CALCULATIONS INVOLVING HEATING AND/OR COOLING AND PHASE CHANGES, THINK ABOUT THE STEPS NEEDED. EACH STRAIGHT LINE STEP NEEDS A SEPARATE CALCULATION.

20. HEAT CAPACITIES FOR WATER IN VARIOUS STATES: Solid Water: C=2.00 J/g/°C Liquid Water: C = 4.18 J/g/°C Gaseous Water: C = 2.01 J/g/°C

21. HOW MUCH ENERGY WOULD BE REQUIRED TO CONVERT WATER AT -20 o C TO WATER AT 120 o C? THINK ABOUT YOUR STEPS.