CIE iGCSE Co-Ordinated Science P1.6.4 Power Exam Style Questions Paper 2
▶️ Answer/Explanation
Total electrical energy input \( = 4000 + 12\,000 = 16\,000 \text{ J} \).
Efficiency \( = \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100 = \dfrac{4000}{16\,000}\times100 \).
This gives an efficiency of 25%.
✅ Answer: (A)
Efficiency \( = \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100 = \dfrac{4000}{16\,000}\times100 \).
This gives an efficiency of 25%.
✅ Answer: (A)
▶️ Answer/Explanation
Power is energy transferred per second: \(P = \frac{E}{t}\).
Converting energy: \(400\,\text{kJ} = 400\,000\,\text{J}\).
\(P = \frac{400\,000}{5.0} = 80\,000\,\text{W}\).
✅ Answer: (C)
Converting energy: \(400\,\text{kJ} = 400\,000\,\text{J}\).
\(P = \frac{400\,000}{5.0} = 80\,000\,\text{W}\).
✅ Answer: (C)
▶️ Answer/Explanation
The work done lifting the lift is calculated using \(W = Fd = 4000 \times 12 = 48000\ \text{J}\).
Power is the rate of doing work, so \(P = \frac{W}{t}\), which rearranges to \(t = \frac{W}{P}\).
Substituting values gives \(t = \frac{48000}{800} = 60\ \text{s}\).
✅ Answer: (D)
Power is the rate of doing work, so \(P = \frac{W}{t}\), which rearranges to \(t = \frac{W}{P}\).
Substituting values gives \(t = \frac{48000}{800} = 60\ \text{s}\).
✅ Answer: (D)
Question
Four different kettles contain different masses of water.
They are used to heat the water from room temperature to boiling point.
The kettles take different times to do this.
Which kettle has the lowest useful power output?
▶️Answer/Explanation
Ans: B
Question
A 900W oven operates for 2.0 minutes. How much energy is transferred by the oven?
(A) 7.5 J (B) 450 J (C) 1.8 kJ (D) 108 kJ
▶Answer/Explanation
Ans : D