1. 16.1: Thermal Energy and Matter

2. Heat Heat is the transfer of thermal energy from one object to another because of a temperature difference. Heat flows spontaneously from hot objects to cold objects.

3. Temperature Temperature is a measure of how hot or cold an object is compared to a reference point. On the Kelvin scale, absolute zero is defined as a temperature of 0 kelvins.

4. Temperature As an object heats up, its particles move faster, on average. The average kinetic energy of the particles increases. One way that heat flows is by the transfer of energy in collisions. On average, high-energy particles lose energy. Low-energy particles gain energy.

5. Thermal Energy Thermal energy is the total potential and kinetic energy of all the particles in an object. Thermal energy depends on the mass, temperature, and phase (solid, liquid, or gas) of an object.

6. Specific Heat The amount of heat needed to raise the temperature of one gram of a material by one degree Celsius. The lower a material’s specific heat, the more its temperature rises when a given amount of energy is absorbed by a given mass.

8. Calculating Specific Heat In this formula, heat is in joules, mass is in grams, specific heat is in J/g°C, and the temperature change is in degrees Celsius.

9. Variables Units Q=Heat Energy Joules (J) m=mass grams (g) c=Specific Heat J/g°C t=change in Temperature °C t final -t initial

10. Steps to Follow to Solve for Heat Energy (Q): 1.If needed, subtract t f -t i to find the change in temperature 2.Multiply m, c & t

11. Steps to Follow to Solve for Temperature (t): 1.Multiply m & c 2.Divide Q by your answer from step 1

12. Steps to Follow to Solve for Specific Heat (c): 1.If needed, subtract t f -t i to find the change in temperature 2.Multiply m & t 3.Divide Q by your answer from step 2

13. Steps to Follow to Solve for Mass (m): 1.If needed, subtract t f -t i to find the change in temperature 2.Multiply c & t 3.Divide Q by your answer from step 2

14. Calculating Specific Heat An iron skillet has a mass of 500.0 grams. The specific heat of iron is 0.449 J/g°C. How much heat must be absorbed to raise the skillet’s temperature by 95.0°C?

15. Calculating Specific Heat An iron skillet has a mass of 500.0 grams. The specific heat of iron is 0.449 J/g°C. How much heat must be absorbed to raise the skillet’s temperature by 95.0°C?

16. Calculating Specific Heat How much heat is needed to raise the temperature of 100.0 g of water by 85.0°C?

17. Calculating Specific Heat How much heat is needed to raise the temperature of 100.0 g of water by 85.0°C? Q = m * c * ∆T = (100.0 g)(4.18 J/g°C)(85.0°C) = 35.5 kJ

18. Calculating Specific Heat How much heat is absorbed by a 750-g iron skillet when its temperature rises from 25°C to 125°C?

19. Calculating Specific Heat How much heat is absorbed by a 750- g iron skillet when its temperature rises from 25°C to 125°C? Q = m * c * ∆T = (750 g)(0.449 J/g°C)(125°C – 25°C) = (750 g)(0.449 J/g°C)(100°C) = 34 kJ

20. Calculating Specific Heat In setting up an aquarium, the heater transfers 1200 kJ of heat to 75,000 g of water. What is the increase in the water’s temperature? (Hint: Rearrange the specific heat formula to solve for ∆T.)

21. Calculating Specific Heat In setting up an aquarium, the heater transfers 1200 kJ of heat to 75,000 g of water. What is the increase in the water’s temperature? (Hint: Rearrange the specific heat formula to solve for ∆T.) ∆T = Q / (m x c) = 1,200,000 J/(75,000 g × 4.18 J/g°C) = 3.8°C

22. Calculating Specific Heat To release a diamond from its setting, a jeweler heats a 10.0-g silver ring by adding 23.5 J of heat. How much does the temperature of the silver increase?

23. Calculating Specific Heat To release a diamond from its setting, a jeweler heats a 10.0-g silver ring by adding 23.5 J of heat. How much does the temperature of the silver increase? ∆T = Q / (m x c) = 23.5 J/(10.0 g × 0.235 J/g°C) = 10.0°C

24. Calculating Specific Heat What mass of water will change its temperature by 3.0°C when 525 J of heat is added to it?

25. Calculating Specific Heat What mass of water will change its temperature by 3.0°C when 525 J of heat is added to it? m = Q / (∆T x c) = 525 J/(3.0°C × 4.18 J/g°C) = 42 g

26. Calculating Specific Heat How many joules of heat are needed to raise the temperature of 10g of aluminum from 22  C to 55  C, if the specific heat of aluminum is 0.90J/g  C?

27. Calculating Specific Heat 100g of 4  C water is heated until its temperature is 37  C. Calculate the amount of heat energy needed to cause this rise in temperature.

28. Calculating Specific Heat How much heat is needed to raise the temperature of 16g of iron from 25  C to 175  C?

29. Calorimeter An instrument used to measure changes in thermal energy.