Let's go through each question in detail.

Question 1

1. Compare the useful output energy by a machine and the input energy supplied to the machine.

   • Answer: The useful output energy of a machine is the portion of the input energy that is converted into the desired form for work. However, some energy is always lost as waste (usually as heat or sound), meaning the output is generally less than the input energy.

2. Explain why the percentage efficiency of a machine can never be:

   • i. more than 100%: It’s impossible for a machine to have more than 100% efficiency because energy cannot be created. The output energy can never exceed the input energy; hence, the machine cannot produce more useful energy than what was initially supplied.
   • ii. equal to 100%: Even with perfect conditions, some energy is always lost due to factors like friction, resistance, or heat. This energy dissipation makes it impossible for a machine to be 100% efficient.

Question 2

An electric motor is used to raise a weight. When you supply 60 J of energy to the motor, the weight gains 24 J of gravitational potential energy.

1. Calculate the energy wasted by the motor.

   • Answer: The energy wasted by the motor is the difference between the input energy and the useful output energy. $\text{Energy wasted} = \text{Input energy} - \text{Useful output energy} = 60 \, \text{J} - 24 \, \text{J} = 36 \, \text{J}$

2. Calculate the efficiency of the motor.

   • Answer: Efficiency is calculated as: $\text{Efficiency} = \frac{\text{Useful output energy}}{\text{Input energy}} \times 100$ Substituting values: $\text{Efficiency} = \frac{24 \, \text{J}}{60 \, \text{J}} \times 100 = 40\%$

Question 3

A machine is 25% efficient. If the total energy supplied to the machine is 3200 J, calculate how much useful energy can be transferred.

• Answer: Since the machine is 25% efficient, the useful energy transferred is: $\text{Useful energy} = \text{Total energy} \times \frac{\text{Efficiency}}{100} = 3200 \, \text{J} \times \frac{25}{100} = 800 \, \text{J}$

Question 4

An electric fan heater contains a motor that blows hot air into the room. Describe the changes in the energy stores of the heater.

• Answer: In an electric fan heater:
  • Electrical energy is first supplied to the heater, which increases the internal energy store of the heater's elements.
  • This internal (thermal) energy is then transferred to the air, increasing the thermal energy store of the air.
  • The fan converts some of the electrical energy into kinetic energy, which helps distribute the warm air throughout the room.
  • Some energy is lost as sound from the fan motor and friction in its moving parts.

Do you want additional details on any part of these answers?

Related Questions:

1. What are some real-world factors that reduce the efficiency of machines?
2. How can the efficiency of an electric motor be improved?
3. Why is energy wasted in the form of heat in most machines?
4. What would happen if a machine could be 100% efficient?
5. How is efficiency calculated differently for various forms of energy, like thermal and mechanical?

Tip:

Always remember, efficiency is a measure of how much input energy is effectively used for the intended output. High efficiency means minimal waste.