Section 7.2—Calorimetry & Heat Capacity

1. Section 7.2—Calorimetry & Heat Capacity Why do some things get hot more quickly than others?

2. Temperature Temperature– proportional to the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As temperature increases Molecules move faster

3. Heat & Enthalpy Heat (q)– The flow of energy from higher temperature particles to lower temperature particles Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”

4. Energy Units The most common energy units are Joules (J) and calories (cal) Energy Equivalents = 4.18 J 1.00 cal = 1000 J 1 kJ = 1000 cal 1 Cal (food calorie) These equivalents can be used in dimensional analysis to convert units

5. Heat Capacity Specific Heat Capacity (C)– The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C Cw (C for liquid water) = 1.00 cal/g°C or 4.18 J/g°C

6. Heat Capacity High Heat Capacity Low Heat Capacity Takes a large amount of energy to noticeably change temp Small amount of energy can noticeably change temperature Heats up slowly Heats up quickly Cools down slowly Cools down quickly Maintains temp better with small condition changes Quickly readjusts to new conditions A pool takes a long time to warm up and remains fairly warm over night. The air warms quickly on a sunny day, but cools quickly at night A cast-iron pan stays hot for a long time after removing from oven. Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

7. What things affect temperature change? • Heat Capacity of substance • The higher the heat capacity, the slower the temperature change • Mass of sample • The larger the mass, the more molecules there are to absorb energy, so the slower the temperature change Specific heat capacity of substance Change in temperature Energy added or removed Mass of sample

8. Positive & Negative DT • Change in temperature (DT) is always final temperature – initial temperature • If temperature increases, DT will be positive • A substance goes from 15°C to 25°C. • 25°C - 15°C = 10°C • This is an increase of 10°C • If temperature decreases, DT will be negative • A substance goes from 50°C to 35°C • 35°C – 50°C = -15°C • This is a decrease of 15°C

9. Positive & Negative DH • Energy must be put in for temperature to increase • A “+” DT will have a “+” DH • Energy must be removed for temperature to decrease • A “-” DT will have a “-” DH

10. Let’s Practice #1 Example: How many joules must be removed from 25 g of water at 75°C to drop the temperature to 30°? Cw = 4.18 J/g°C

11. Let’s Practice #1 Example: How many joules must be removed from 25.0 g of water at 75.0°C to drop the temperature to 30.0°? Cp water = 4.18 J/g°C DH = change in energy m = mass Cw = heat capacity of water DT = change in temperature (Tf - Ti) DH = - 4703J

12. Let’s Practice #2 If the specific heat capacity of aluminum is 0.900 J/g°C, how much heat must be added to a 30.0 g sample at 15°C to increase its temperature to 30°C ?

13. Calorimetry

14. Conservation of Energy 1st Law of Thermodynamics– Energy cannot be created nor destroyed in physical or chemical changes This is also referred to as the Law of Conservation of Energy If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

15. Calorimetry Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system Because of the Law of Conservation of Energy, the energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.

16. Calorimetry DHsurroundings = - DHsystem (m×C×DT)surroundings = - (m×C×DT)system

17. Calorimetry Calorimetry also uses the principle of thermal equilibrium. Thermal Equilibrium– Two objects at different temperatures placed together will come to the same temperature

18. An example of Calorimetry Example: A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the specific capacity of the metal?

19. An example of Calorimetry Example: A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal? Step 1: Find the Heat (ΔH) of the water ΔH = mCwΔT ΔH = (50)(4.184)(32.5 – 24) ΔH = 1778.2 J Step 2: Remember that the water is the “surroundings”. So, the heat gained by the water is the same as the heat lost by the metal but opposite in sign!! Metal: m = 23.8 g Ti = 100.0°C Tf = 32.5°C Cm = ? Water: m = 50.0 g Ti = 24°C Tf = 32.5°C Cw = 4.184 J/g°C

20. Step 3: Since you know ΔH of the metal system (-1778.2 J), use this the find Cm. ΔH = mCmΔT Now use the values for the metal! -1778.2 = (23.8)Cm(32.5 – 100) -1778.2 = (23.8)(-67.5)Cm (23.8)(-67.5) Cm = 1.11 J/goC

21. Let’s Practice Example: A 10.0 g of aluminum at 95.0°C is placed in a container of 100.0 g of water (C = 4.184 J/g°C) at 25.0°. The metal and water reach a final temperature of 26.5 oC. What is the specific heat of the metal?