Specific Heat Capacity (比熱容量)

temperature T Energy Q

3

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

(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.

Different materials have different values of specific heat capacities: 6

7

8

0.284 28 35 7 99700 109500 9800 4930

12

higher 13

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

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

Class Exercise 10.1 16

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

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

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 = 512.5 J / kg C 20

Class Exercise 10.2

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

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 = 24 + 2 = 26 C 23

It is because heat loss to surroundings. 25

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

Class Exercise 10.3

28

Q = 14005 – 12345 = 1660 J 29

Assume no energy loss to surroundings. Q = 14005 – 12345 = 1660 J 1660 J Assume no energy loss to surroundings. 30

31

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

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

Class Exercise 10.4

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

36

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

 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

Class Exercise 10.5 39

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

Class Exercise 10.5 41

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

Class Exercise 10.5 43

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

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