1. Specific Heat Capacity (比熱容量)

2. temperature T
Energy Q

3. 3

4. Q energy [J]
m mass [kg]
T temperature change [C]
c specific heat capacity [J / kg C]

5. (e) In words, specific heat capacity of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 C.
The unit of specific heat capacity is J / kg C.
It is found that each substance has a unique value of specific heat capacity.

6. Different materials have different values of specific heat capacities:6

7. 7

8. 8

11. 0.284 28 35 7
99700
109500
9800
4930

12. 12

13. higher 13

14. Heat loss to surroundings
higher
Heat loss to surroundings
Energy absorbed by polystyrene cup, stirrer, … 14

15. Heat loss to surroundings
higher
Heat loss to surroundings
Energy absorbed by polystyrene cup, stirrer, …
It is because the heater still releases energy to the water and the temperature is rising. 15

16. Class Exercise 10.1 16

18. Q = mcT
35 = (3.81000)(c)(18-10)
c = 1150 J / kg C 18

19. 35 = (3.81000)(c)(18-10) c = 1150 J / kg C 650 = (1)(c)(25-20)
Q = mcT
35 = (3.81000)(c)(18-10)
c = 1150 J / kg C
Q = mcT
650 = (1)(c)(25-20)
c = 130 J / kg C 19

20. 35 = (3.81000)(c)(18-10) c = 1150 J / kg C 650 = (1)(c)(25-20)
Q = mcT
35 = (3.81000)(c)(18-10)
c = 1150 J / kg C
Q = mcT
650 = (1)(c)(25-20)
c = 130 J / kg C
Q = mcT
4100 = (2)(c)(26-22)
c = J / kg C 20

21. Class Exercise 10.2

22. Class Exercise 10.2 Q = mcT 12600 = (0.5)(c)(24-18)
c = 4200 J / kg C 22

23. Final temperature = 24 + 2 = 26 C
Class Exercise 10.2
Q = mcT
12600 = (0.5)(c)(24-18)
c = 4200 J / kg C
T = 8400  4200 = 2
Final temperature = = 26 C 23

25. It is because heat loss to surroundings.25

26. It is because heat loss to surroundings.
It is the room temperature. 26

27. Class Exercise 10.3

28. 28

29. Q = – 12345
= 1660 J 29

30. Assume no energy loss to surroundings.
Q = – 12345
= 1660 J
1660 J
Assume no energy loss to surroundings. 30

31. 31

32. Q = mcT
1660 = (0.5)(c)(28-20)
c = 415 J / kg C 32

33. To reduce the heat conducted through the table.
Q = mcT
1660 = (0.5)(c)(28-20)
c = 415 J / kg C
To reduce the heat conducted through the table. 33

34. Class Exercise 10.4

35. Q = mcT
30 = (0.0283)(c)(8)
c = 133 J / kg C 35

36. 36

37. Q = mcT
40 = (0.0283)(c)(6)
c = 236 J / kg C 37

38.  It is not a gold or silver coin.
Q = mcT
40 = (0.0283)(c)(6)
c = 236 J / kg C
Q = mcT
110 = (0.0283)(c)(10)
c = 389 J / kg C
 It is not a gold or silver coin. 38

39. Class Exercise 10.5 39

40. Class Exercise 10.5
Q = (60)(1)(3600) = 216,000 J 40

41. Class Exercise 10.5 41

42. Class Exercise 10.5
Q = mcT
= (2) (4200) (45-20)
= 210,000 J 42

43. Class Exercise 10.5 43

44. Class Exercise 10.5
Q = mcT
216,000 – 210,000 = (0.5) (c) (45-20)
6000 = 12.5 c
c = 480 J / kg C 44

45. Q = mcT
216,000 – 210,000 = (0.5) (c) (45-20)
6000 = 12.5 c
c = 480 J / kg C
The calculated value will be smaller than the value in (i) because less energy will be absorbed by the metal and the temperature rise will be smaller. 45