3.4 
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1 - Science of Physics
2 - Forces and Motion3 - Energy Transformations      Chapter 3 at a glance      3.1 - Energy      3.2 - Conservation of energy      3.3 - Flow of energy      3.4 - Temperature and the kinetic theory of matter      3.5 - Chapter review4 - Waves and Sound
5 - Electricity and Magnetism
6 - Light and Optics
7 - The Atom
8 - Linear Motion
9 - Vectors and Forces
10 - Forces
11 - Newton's Laws and Gravitation
12 - Circular Motion and Gravitation
13 - Torque and Static Equilibrium
14 - Momentum and Collisions
15 - Angular momentum
16 - Power and Machines
17 - Harmonic Motion
18 - Waves
19 - Sound
20 - Electricity
21 - Electric and Magnetic Fields
22 - Electromagnetism and Induction
23 - Light and Color
24 - Geometrical Optics
25 - Electromagnetic Radiation
26 - The Properties of Matter
27 - Thermodynamics and Heat Transfer
28 - The Atom
29 - The Quantum Theory
30 - TBD
31 - TBD
32 - TBD
33 - TBD
34 - TBD
35 - TBD
36 - TBD
37 - The atomic nucleus and radioactivity
38 - Nuclear reactions
39 - Relativity
40 - Particle physics and the standard model
41 - Electronics (semiconductors)
42 - Nuclear technology
43 - Lasers and photonics
44 - Metals, alloys and materials science
45 - Communications technology
46 - Buildings and structures
47 - Energy technology
48 - Biophysics
49 - Matter and Energy, Space and Time
50 - Energy Transformations
51 - Nanotechnology
52 - Website Videos
53 - ExtraStuff
55 - Modeling Motion, Friction, and Center of Mass
56 - Work and the Conservation of Energy
Specific heat
Thermal energy From experience you probably know that an object has more thermal energy when it has a higher temperature compared to when it has a lower temperature. The amount of thermal energy in a certain quantity of matter, however, also depends other factors such as mass, composition, and whether the matter is solid, liquid, or gas.
Specific heat Thermal energy and specific heat Different materials contain different amounts of thermal energy, even when they are at the same temperature. It takes 4.18 J of energy to raise the temperature of 1 g of water by 1°C, while it only takes 0.9 J to raise the temperature of 1 g of aluminum by 1°C. The amount of thermal energy per degree per gram is called the specific heat. Water has a specific heat of 4.18 J g−1 °C−1 and that means water contains 4.18 joules of thermal energy per degree per gram. Aluminum has a specific heat of only 0.9 J g−1 °C−1 and therefore aluminum only contains 0.9 joules of thermal energy per degree per gram.
Equation for specific heat The thermal energy of an object depends on both its temperature and the specific heat of its material. As a practical matter, we usually measure an object's thermal energy relative to another one—since a temperature difference between two objects will cause heat to flow from one to the other. In calculating the object's thermal energy, we therefore use its relative temperature compared to another object, ΔT=T1−T2.
Specific heat equation
(3.13) E th = m c p ( T 2 − T 1 ) = m c p ΔT
Eth  = thermal energy or heat (J)
m  = mass (kg)
cp  = specific heat capacity (J kg-1 °C-1)
T1  = temperature number one (K or °C)
T2  = temperature number two (K or °C)
ΔT  = change in temperature (K or °C) = T1-T2
Specific heat
Test your knowledge How much heat does it take to raise the temperature of 3 kg of aluminum by 10° C?
  1. 9,000 J
  2. 27,000 J
  3. 480,300 J
  4. 27,000,000 J
Show
Solved Problem 3.9: Energy required to heat water
What is the energy required to heat 100 g of water from 30°C to 40°C? (Specific heat of liquid water is 4,186 J kg−1 °C−1.)

Asked: Thermal energy required to heat the water
Given: Initial temperature T2=30°C
Final temperature T1=40°C
Mass of the water m=100 g
Specific heat of water cp=4,186 J kg−1 °C−1
Relationships: Thermal energy equation as a function of a material's specific heat:
E th = m c p ( T 2 − T 1 ) = m c p ΔT
Solution: The mass of water is in grams and first must be converted to kg to match the units of specific heat constant:
m=( 100 g )( 1 kg 1000 g )=0.1 kg
Now we substitute values into the equation to get the energy required:
E th =m c p ( T 1 − T 2 )=(0.1 kg)(4,186  J kg −1  ° C −1 )(40°C−30°C)=4,186 J
Answer: 4,186 J

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