Thermal Properties of Matter (Part II)

1. Thermal Properties of Matter (Part II)

2. Topics Internal Energy Specific Heat Capacity Specific Latent Heat Combined Problems

3. Internal Energy • Previously, we have talked about the difference between Thermal Energy and Heat. • Recall that heat is the thermal energy when it is moving (usually hotter to colder region) • Internal energy refers to the thermal energy within a substance when there is no heat flowing • Internal energy is dependent on 2 things: • number of molecules • Temperature of substance

4. Internal Energy Which has more internal energy, a human being at temperature of 37°C or the ocean at 20°C? Ans: the ocean, since it has much more molecules than the human, even though the average KE of molecules in the human is greater than the average KE of molecules in ocean Qn: when the human jumps into the ocean, where does heat flow? Ans: heat flows from higher to lower temperature, NOT from higher to lower internal energy!!

5. Specific Heat Capacity Suppose I use a bunsen burner to heat up 1 kg piece of metal, and I use the same bunsen burner to heat up 1 kg of water. Which one would heat up to 100°C faster?? Ans: the metal. Why? The metal requires less heat (thermal energy) to increase its temperature, while water requires more heat to increase the same amount of temperature.

6. Specific Heat Capacity We say that water has a higher specific heat capacity than metal Specific heat capacity is defined as the amount of thermal energy required to raise the temperature of a unit mass of a substance by 1 K or 1° Symbol: small case “c” Units: Jkg-1K-1 OR Jg-1K-1

7. Specific Heat Capacity Analogy: Macarius only need to eat 1 plate of rice to feel full, Alex needs to eat 2 plates of rice to feel equally full. After they both have eaten, even though they are equally fully, Alex needs twice as much food to feel just as full. Similarly, if material X has twice as much specific heat capacity as material Y, 1 kg of X would require twice as much thermal energy (heat) to increase its temperature by the same amount compared to 1 kg of Y

8. Specific Heat Capacity • Equation to memorize: • Q = mcθ • Q is heat supplied or given out, in Joules • m is mass of substance, in kg (or g) • c is specific heat capacity, in Jkg-1K-1 (or Jg-1K-1) • θ is change in temperature, in °C or K • Protip: use this formula to remember units of c

9. Worked Example 1 • Water has specific heat capacity of 4200 Jkg-1K-1 . If 5000 J of heat was supplied to 1 kg of water, determine the increase in temperature of the water. • Q = mcθ • 5000 = (1)(4200)θ • θ= 1.19 °C (3 sf) • The water increased its temperature by 1.19 °C

10. Worked Example 2a A cup of containing 200g of boiling water was allowed to cool from 100°C to 25°C. If the specific heat capacity of water is 4200 Jkg-1K-1, determine how much heat was given out by the water if the water is in thermal equilibrium with the cup.

11. Worked Example 2b If the mass of the cup is 300 g and the specific heat capacity of the cup is 2.00 Jg-1K-1 , determine the total amount of heat given off by the cup AND water.

12. Strategy for solving more complex problems • When there are two or more substances interacting (i.e, passing heat from one to another), • Step 1: treat the two substances separately in your workingto determine Q for each substance • Step 2: Heat gain by one substance = heat loss by other substance • If temperature is an unknown, use algebra to solve (e.g. Let X be the final Temperature)

13. Worked Example 3 200 g of cold water at 10 °C is mixed with 300g of warm water at 50 °C. What is the resulting temperature of water, if the specific heat capacity of water is 4200 Jkg-1K-1 ? Assume no heat loss to surroundings.

14. Worked Example 4 A 150 g piece of metal was heated to 200 °C, and then placed in 500 g of water, initially at 25 °C. Determine the final temperature when the metal and water are in thermal equilibrium. Assume no heat loss to surroundings. The Specific heat capacities of metal and water are 1000 Jkg-1K-1 and 4200 Jkg-1K-1respectively.

15. Heat Capacity • Heat Capacity • not to be confused with specific heat capacity • Symbol: capital letter “C” • Units: JK-1 • relationship to c: C = mc • m is mass • c is specific heat capacity

16. Heat Capacity Definition: Heat Capacity C is the amount of thermal energy required to raise the temperature of a substance by 1 K (or 1 C) Note: Heat Capacity (and its definition) is in syllabus, but rarely tested. Specific Heat Capacity is more commonly tested.

17. Specific Latent Heat Just as specific heat capacity is the heat required to increase the temperature of 1 kg of substance, specific latent heat is the heat required to change the state of 1 kg of substance Symbol: lower case l Units: J kg-1 Equation to memorize: Q = ml

18. Specific Latent Heat There are two kinds of specific latent heat. Specific latent heat of fusionlf of a substance is the amount of thermal energy required to change unit mass of the substance from solid state to liquid state, without a change in temperature Specific latent heat of vaporisationlvof a substance is the thermal energy required to change unit mass of the substance from liquid state to gaseous state, without a change in temperature

19. Worked Example 5 The specific latent heat of vaporisation of water is 2200 kJ kg-1. Determine how much water is converted to steam when 1000 kJ of energy is supplied to water at 100 °C.

20. Worked Example 6 51 kJ of energy is released when 150 g of substance X solidifies at melting point, determine the specific latent heat of fusion of X.

21. Latent Heat • Latent Heat • Not to be confused with specific latent heat. • Symbol: Capital letter “L” • Units: J • Relationship to l: • L = ml

22. Latent Heat Latent heat of fusion Lf is the amount of thermal energy required to change a substance from solid state to liquid state, without a change in temperature Latent heat of vaporisationLv is the amount of thermal energy required to change a substance from liquid state to gaseous state, without a change in temperature.

23. Summary of 4 quantities

24. Strategy for solving combined problems If there is more than one substanceinteracting, consider them separately. If one substance undergoes more than one change, break it apart into different phases and consider them separately (e.g. Phase 1 – increase of temperature to boiling point, Phase 2 – conversion from liquid to gas) Use algebra to solve for unknowns (e.g. let X be final temperature, etc.)

25. Worked Example 7 Specific heat capacity of water is 4200 Jkg-1K-1 and the specific latent heat of vaporisation is 2200 kJ kg-1. Determine how much energy it would need to convert 500g of water at 25 °C completely to steam.

26. Worked Example 8 Specific heat capacity of water is 4200 Jkg-1K-1 and the specific latent heat of vaporisation is 2200 kJ kg-1. 50 g of steam at 100 °C was pumped into 500g of water at 25 °C. If all the steam was condensed into water, determine the final temperature of water. Ans: 71.3 °C

27. Summary • 6 definitions: Specific Heat Capacity, Heat Capacity, Specific Latent Heat, Latent Heat • 2 Equations • Q = mcθ • Q = ml • Solving quantitative problems which include either or both of these two equations

28. 15 min Quiz!!