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SCIENCE
Module 3
GRADE 8
Prepared by: Engr. L.N. Abrigo

MODULE 3: HEAT AND TEMPERATURE
Lesson 1 – HEAT VS. TEMPERATURE
WHAT IS HEAT?
● Heat is viewed as a form of energy that is
transferred from one body due to a difference in
temperature.
● The SI unit of heat is Joule (J).
● A more common unit is calorie (cal) which is
defined as the amount of heat needed to change
the temperature of one gram of water by 1 degree
celsius ( )

WHAT IS TEMPERATURE?
● Temperature is the “hotness” or coldness” of a
substance.
● It is the measure of the motion of the molecules or
atoms within the substance (Particle Theory of
Matter).
● Thus, on the molecular level, temperature is the
measure of the average kinetic energy of the

Lesson 2 – HEAT EFFECTS ON MATTER (Phase
Change)
PHASES OF MATTER

Lesson 2 – EFFECTS ON MATTER (Phase Change)

Lesson 3 – HEAT CAPACITY
WHAT IS HEAT CAPACITY?
● the amount of heat needed by a material to
increase its temperature by a degree
Note: What is Thermal/Heat Energy?
-its the energy that is actually contained in an
object due to the motion of its particles

Lesson 3 – HEAT CAPACITY
Q = Heat energy
m = mass
c = specific heat
ΔT = change in temperature
= (Tfinal – Tinitial)

Lesson 3 – HEAT CAPACITY (SAMPLE PROBLEM)
Question 1: A 500 gram cube of lead is heated from 25 °C to 75 °C. How much
energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.
Solution: First, let’s the variables we know.
m = 500 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C
Plug these values into the specific heat equation from above.
Q = mcΔT
Q = (500 grams)·(0.129 J/g°C)·(50 °C)
Q = 3225 J
Answer: It took 3225 Joules of energy to heat the lead cube from 25 °C to 75 °C.

Lesson 3 – HEAT CAPACITY (SAMPLE PROBLEM)
Question 2: A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of
the metal?
Solution: List the information we know.
m = 25 grams
ΔT = 200 °C
Q = 2330 J
Place these into the specific heat equation.
Q = mcΔT
2330 J = (25 g)c(200 °C)
2330 J = (5000 g°C)c
Divide both sides by 5000 g°C
c = 0.466 J/g°C
Answer: The specific heat of the metal is 0.466 J/g°C.

Lesson 4 – TEMPERATURE CONVERSION

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Grade 8 Science Module 3

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