Specific Heat

# Specific Heat

## Specific Heat

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##### Presentation Transcript

1. Specific Heat

2. Different substances have different abilities to store energy. They are said to have different heat capacities. • Heat capacity is defined as the amount of heat required to change the temperature of the substance by one degree Celsius (or Kelvin).

3. Most metals have low specific heats, while nonmetal compounds & mixtures such as water, wood, soil, & air have relatively high specific heats.

4. Calorimeter • A device that measures temperature changes in surroundings • Heat transferred by physical and chemical changes can be measured using a process called calorimetry.

5. Chem Saver p 41CALORIMETRY • Calorimetry: Process for determining the amount of heat energy released or absorbed in a chemical or physical change. • Enthalpy : Change in Heat energy Symbolized by H • Entropy : extensive property which measures the degree of disorder.

6. Units of heat include: • calorie - the amount of heat required to change the temperature of 1 gram of pure liquid water by one degree Celsius. • Food calorie (Calorie, big calorie) - is equal to 1000 calories or one kilocalorie • Joule - International System unit of energy. There are 4.18 Joules in one calorie. • Kilojoule - 1000 joules • For heat conversions use: • 1Kcalorie= 1Calorie = 1000 calories • 1calorie = 4.18J

7. Foods and Heat Energy • 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. • Energy in our bodies comes from carbohydrates and fats (mostly). • Intestines: carbohydrates converted into glucose: • C6H12O6 + 6O2 6CO2 + 6H2O, DH = -2816 kJ • Fats break down as follows: • 2C57H110O6 + 163O2 114CO2 + 110H2O, DH = -75,520 kJ • Fats contain more energy; are not water soluble, so are good for energy storage.

8. Problem #1 A popscicle has 60.0 Calories per serving. How many calories is this? • 1Kcalorie= 1Calorie = 1000 calories • 1calorie = 4.18J • 60.0Calories x 1000calories =60,000 cal or 1 1Cal 6.00x104cal

9. Problem #2 Amazing fruit candy has 10.0 Calories per serving. How many joules is this? • 1Kcalorie= 1Calorie = 1000 calories • 1calorie = 4.18J • 10.0 Calories x 1000 cal x 4.18J = 41800J 1 1Cal 1cal

10. Chem Saver Page 41Calculating Heat Energy Since the calorie is defined in terms of water, the heat capacity for liquid water is 1 cal/g oC. • This also equates to 4.18 J/g oC. • Calculating calorimetry problems; • Q = m x Cp x ΔT • Q = heat (cal or J) • m = mass of the substance (g) • Cp = heat capacity (cal/g oC or J/g oC) • and D T = change in temperature of the substance (oC)

11. The greater the mass of the object, the greater its heat capacity. A massive steel cable on a bridge requires much more heat to raise its temperature 1ºC than a small steel nail does. Different substances with the samemass may have different heat capacities. On a sunny day, a 20-kg puddle of water may be cool, while a nearby 20-kg iron sewer cover may be too hot to touch. Heat Capacity

12. Problem #7 When a hamburger is burned in a calorimeter, 2000. g of water increases in temperature by 30.0 oC. How many Calories are in the hamburger? Q=(m) (Cp) (T) • Cwater = 1 cal/g oC or 4.18 J/g oC • T = 30 C • Q = 2000. g x 1 cal/g oC x 30.0 oC = 60000cal • 60000cal x 1Cal = 60.0 Cal 1 1000cal

13. Problem #9 The temperature of 2500 grams of mercury rises from 20 oC to 60oC when it absorbs 13,794 joules of heat. Calculate the specific heat capacity of the mercury. Q=(m) (Cp) (T) • Cmercury =? • ΔT = 40 oC • 13,794 J = 2500g x (Cp) x 40.0 oC • 13,794 J = 100000g oC x (Cmercury) • (Cmercury)=0.138 J/g oC

14. Problem #11a • An 800-gram block of lead is heated in boiling water (100 oC) until its temperature is the same as the boiling water. The lead is then removed from the boiling water and dropped into 250 grams of cool water at 12.2 oC. After a short time, the temperatures of both lead and water levels out at 20.0 oC. • Calculate the amount of heat (in Joules) gained by the cool water. • Q=(m) (Cp) (T) m= 250g Ti= 12.2 C and Tf= 20.0 C T = 7.8 C Cwater= 4.18 J/gC Q = (250g) ( 4.18 J/gC)( 7.8 C ) Q =8151J

15. Problem #11b • An 800-gram block of lead is heated in boiling water (100 oC) until its temperature is the same as the boiling water. The lead is then removed from the boiling water and dropped into 250 grams of cool water at 12.2 oC. After a short time, the temperatures of both lead and water levels out at 20.0 oC. Calculate the specific heat capacity of the lead based on these measurements, assuming that no heat was lost in the process. • Q=(m) (Cp) (T) m= 800g T = 80 C CPb= ? Q gained by water= Q lost by Pb = 8151J 8151J = (800g) (CPb) (80 C) CPb = 0.127 J/gC

16. At the freezing or boiling point two phases of matter can exist at the same temperature To make the change from one phase to another more energy will be absorbed (boiling or melting) or lost (condensing or freezing) without a change in temperature This is because this energy is used merely to overcome the bonds of one state and move to the new state creating a change in potential energy. Energy and Change of State

17. Chem Saver page 16Heating and Cooling Curve Heat of vaporization Gas/vapor D PE Boiling/vaporization D KE condensation Heat of fusion D PE liquid melting D KE D KE Solid freezing 100oC 0oC Time

18. Instead of specific heat (C) we use enthalpy (∆H) for calculating heat during phase changes. Heat of fusion/solidification is the heat required to move from solid  liquid Hfus = –Hsolid Heat of vaporization/condensation is the heat required to move from liquid gas Hvap = –Hcon Heat in phase changes

19. Q = m x H Heat in phase changes • Q = amount of heat energy (joules or calories) • m = mass of substance (grams) • H = enthalpy of fusion (Hf) or vaporization (Hv)

20. Chemistry in ActivityChem Saver Page 41 • constants: Specific heat of ice = 2.09 J/g·ºC Specific heat of water = 4.18 J/g·ºC Specific heat of steam = 2.03 J/g·ºC Heat of fusion of water = 334 J/g Heat of vaporization of water = 540 J/g

21. Example Problem • Calculate the mass of ice (in grams) that will melt at 0ºC if 2.25 kJ of heat are added. (Hf= 334 J/g) Q = m x Hf • Q = 2250 J • Hf = 334 J/g • m = Q /Hf • m = 2250 J / 334 J/g m = 6.74 g

22. Example Problem • Calculate the mass of water vapor (in grams) at 100ºC that can be condensed into liquid at 100ºC if 55.0 kJ of heat is removed. (Hv = 2257 J/g) Q = m x Hv • Q = 55000 J • Hv= 2257 J/g m • m = Q /Hv • m = 55000 J / 2257 J/g • m = 24.4 g

23. Sample Problem • How much heat does it take to turn a 20 g chunk of ice at -40oC into 20 g of steam at 120oC? • This is a 5 step problem, each segment of the graph must be calculated separately and then added together to get a total heat absorbed.

24. Step 1 -40oC

25. Step 1 • How much heat does it take to turn a 20 g chunk of ice at -40oC into 20 g of ice at 0oC? Q=(m) (Cp) (T) m= 20g Ti= -40 C and Tf= 0 C T = Tf - Ti =0 C – (-40.0 C)= 40 C Cice= 2.09 J/g·ºC Q = (20g) ( 2.09 J/gC)( 40 C ) Q = 1672J (+Q indicates heat gained or endothermic)

26. Step 2 • How much heat does it take to turn a 20 g chunk of ice at 0oC into 20 g of water at 0oC? Q=(m) (Hf) m= 20g Hf= 334 J/g Q = (20g) ( 334J/g) Q = 6680J (+Q indicates heat gained or endothermic)

27. Step 3 • How much heat does it take to turn a 20 g water at 0oC into 20 g of water at 100oC? Q=(m) (Cp) (T) m= 20g Ti= 0 C and Tf= 100 C T = Tf - Ti = 100 C – 0C= 100 C Cwater= 4.18 J/g·ºC Q = (20g) ( 4.18 J/gC)( 100 C ) Q = 8360J (+Q indicates heat gained or endothermic)

28. Step 4 • How much heat does it take to turn a 20 g water at 100oC into 20 g of water vapor at 100oC? Q=(m) (Hv) m= 20g Hf= 334 J/g Q = (20g) ( 334J/g) Q = 6680J (+Q indicates heat gained or endothermic)

29. Entropy • Entropy is a measure of how chaotic a system is. The lessorder that is present the more entropy. In terms of states of matter: Liquid Gas Solid Low Entropy High Entropy