LEARNING WORKSHEET
Quarter 1 Science Grade 8
Worksheet No. _____
MELCS: S8FE-If-27
Name: ________________________________
Section: ______________________________
Score: _______________________
Date Submitted: _______________
Heat and Temperature
ACTIVITY 1 – THE EFFECT OF TEMPERATURE ON THE DISPERSION OF
MOLECULES
Objectives:
 Demonstrate how solid particles react in different temperatures of water.
 Compare the speed of dispersion of solid particles in the three setups.
 Calculate the heat absorbed or released to raise or decrease the
temperature of a substance.
Note: Perform this Home-based activity to understand our lesson better. If the
specific materials listed in the activity are unavailable, you may improvise on
whatever material is available at home.
Materials:
 Three jars or three transparent cups
 1 ½ cup of tap water (room temperature)
 1 ½ cold water
 1 ½ hot water
 Three teaspoons of coffee powder
 Timer
Procedure:
1. Label jars/cups A, B, and C, respectively.
A
B
C
Figure 1. Jars labeled A, B, and C.
2.
3.
4.
5.
6.
Fill Jar A with tap water, B with cold water, and C with hot water.
Put a teaspoon of coffee in the center of each jar.
Record the time of the dispersion of the coffee particles using a timer.
Observe the behavior of the coffee powder in the three setups.
Record your observations in the table below.
Setup
Type of Water
A
B
C
Tap
Cold
Hot
Time of
Dispersal
Observations
Guide Questions:
1. What similarities did you observe when you added the coffee powder to
1 Heat and Temperature
2022, Delia B. Bernadas
2.
3.
4.
5.
each container?
_______________________________________________________________________
In which container did the coffee scatter the fastest? In which container
did it scatter the slowest?
_______________________________________________________________________
In which container did the water molecules contain more kinetic energy?
In which container did the water molecules contain the least kinetic
energy?
_______________________________________________________________________
How do you relate the water temperature at the time of the scattering of
the coffee powder?
_______________________________________________________________________
How is temperature related to the speed of particle dispersion?
_______________________________________________________________________
READ THIS:
Application: Specific Heat Capacity
Take the case of copper–bottom pans and Pyrex glass used in the kitchen.
When cooking, copper–bottom pans are the best choice due to their low specific
heat capacity, while Pyrex makes better baking wares due to their high specific
heat capacity.
Specific heat capacity is the amount of heat that will change the
temperature of an object a degree higher or lower. Copper–bottom pans require
a small amount of heat to change their temperature, making them good
cookware, while Pyrex glass requires a large amount of heat to change its
temperature, making it suitable for baking.
Different materials have different specific heat capacities. Many metals
have low specific heat capacities, making them easy to heat up and cool down.
On the other hand, water has a high specific heat capacity, so it takes a long
time to heat water and a long time to cool down. This characteristic makes water
a suitable coolant for car radiators. Water can absorb a large amount of heat
without causing its temperature to rise too high.
The unit of specific heat capacity is J/kg°C, with its mathematical
expression:
𝐽
𝑄
𝐶=
=
𝑘𝑔°𝐶
𝑚∆𝑇
Alternatively, the relation among these quantities is expressed as:
𝑄 = 𝑚𝑐∆𝑇
Where:
 Q = heat (Joules, J)
 ∆T = change in temperature (°C)
 m = mass (kg)
 c = specific heat (J/kg°C)
2 Heat and Temperature
2022, Delia B. Bernadas
How to derive the formula using the Triangular Method:
Q
m c ∆𝐓
Q = mc∆T
c=
𝑄
𝑚∆𝑇
m=
𝑄
𝑐∆𝑇
∆𝑇 =
𝑄
𝑀𝐶
∆𝑇 = 𝑇𝑓 − 𝑇𝑖
Sample Problem:
Solve the problem using the GUESA method.
A 1.50 kg tap water heats from an initial temperature of 30°C to a final
temperature of 80°C. How much heat must be added to the water?
Given: m = 1.50 kg
Ti = 30°C
Tf = 80°C
c = 4186 J/kg°C (specific heat of water)
Unknown: Q =?
Equation:
Q = mc∆T
Solution: Q = (1.50 kg) (4186 J/kg°C) (Tf – Ti)
Q = (1.50 kg) (4186 J/kg°C) (80°C – 30°C)
Q = (6279 J/°C) (50°C)
Answer: Q = 314,000 J
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3 Heat and Temperature
2022, Delia B. Bernadas
ACTIVITY 2 – CALCULATING HEAT AND TEMPERATURE
Solve: (Use the GUESA method.)
1. A piece of unknown metal with a mass of 0.157 kg loses 1090 J heat when
cooled from 175°C to 25°C. What is the specific heat capacity value of this
metal?
2. The temperature of a sample of water increases from 20 0C to 46.6 0C as
it absorbs 56.50 calories of heat. What is the mass of the sample?
3. Calculate the temperature change when:
a. 10.0 kg ow water loses 232kJ of heat. (CW = 4.179J/g 0C)
b. 1.96kJof heat is added t0 500g of copper. (CCu=0.385J/g 0C)
References:
Chalfant, H. et.al. (2005, Fall). Heat and temperature. Sci Ed, 491. Retrieved
from http://www.biol.wwu.edu/donovan/SciEd491/HeatTempUnit.pdf
Expansion and contraction. (n. d.) Retrieved from
http://schools.cbe.ab.ca/b682/pdfs/Science%207/Heat-and
Temperature-Unit3_T4_T6.pdf
4 Heat and Temperature
2022, Delia B. Bernadas
Alumaga et al. (2014, Vibal). Science and Technology, 56-63.
5 Heat and Temperature
2022, Delia B. Bernadas