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Solving Exercise C1 on page 45 Physics 9 textbook
Problem:
Calculate the electrical energy A of the current passing through the resistive wire in that time.
Solution:
Solving Exercise C2 on page 45 Physics 9 textbook
Problem:
Calculate the heat energy Q received by water and an aluminum container in that time.
Solution:
Solving Exercise C3 on page 45 Physics 9 textbook
Problem:
Compare A with Q and provide comments, noting that a small portion of heat is transferred to the surrounding environment.
Solution:
+ Comparison: we observe that A is slightly greater than Q. The consumed electrical energy has transformed into a small amount of heat transferred to the surrounding environment.
Solving Exercise C4 on page 45 Physics 9 textbook
Problem:
Explain the phenomenon stated in the opening section: Why does the hair filament of a light bulb heat up to a high temperature with the same current passing through, while the wire connected to the bulb hardly heats up?
Solution:
The current flowing through both the hair filament and the connecting wire has the same intensity as they are connected in series. According to Ohm's Law, the heat produced in the hair filament and the connecting wire is proportional to their respective resistances. The hair filament has a high resistance, leading to more heat dissipation, resulting in the hair filament heating up to a high temperature and emitting light. On the other hand, the connecting wire has a low resistance, causing minimal heat dissipation, with most of it transferred to the surrounding environment. Therefore, the connecting wire hardly heats up and remains at a temperature close to the ambient temperature.
Solving Exercise C5 on page 45 of Physics Grade 9 Textbook
Problem:
An electric kettle labeled 220V - 1,000W is used with a voltage of 220V to boil 2 liters of water starting from an initial temperature of 20°C. Disregarding heat loss to the kettle's surroundings and heat dissipation, calculate the boiling time for the water. Given the specific heat capacity of water as 4,200 J/kg.K.
Solution:
The electric kettle operates at its rated voltage, so its power (P) is equal to the rated power (1000W).
Neglecting the heat required to warm the kettle's casing and heat dissipation to the surroundings, the heat (Q) needed to boil the water is precisely equal to the electrical energy (A) consumed by the kettle.
Understanding the relationship between electrical energy and power in a current is a crucial lesson in Chapter I - Electricity. Refer to exercise hints on pages 37, 38, 39 of Physics Grade 9 to enhance your understanding.
Applying the Joule-Lenz law is the next topic in Chapter I - Electricity of Physics Grade 9. Check exercise hints on pages 47, 48 for a solid grasp of the concepts and to excel in Physics Grade 9.
