Case Study Practice for KS3 CIE Physics | KS3 CIE 物理：案例分析实战演练

📚 Case Study Practice for KS3 CIE Physics | KS3 CIE 物理：案例分析实战演练

Case studies are a powerful way to link physics principles with real-world situations. In KS3 CIE Physics, you are often asked to interpret data, perform calculations, and explain phenomena using scientific reasoning. This article provides a series of practical case study exercises, covering key topics such as energy, electricity, forces, density, waves, and thermal physics. Each case is presented with a scenario, followed by guided analysis and step-by-step solutions. By working through these examples, you will strengthen your problem-solving skills and deepen your understanding of how physics applies to everyday life.

案例分析是将物理原理与现实情境联系起来的有效方法。在 KS3 CIE 物理中，你经常需要解释数据、进行计算并运用科学推理解释现象。本文提供了一系列实用的案例分析练习，涵盖能量、电学、力、密度、波和热物理等关键主题。每个案例都提供了场景，然后是引导性分析和逐步解答。通过这些练习，你将增强解决问题的能力，加深对物理学如何应用于日常生活的理解。

1. The Electric Kettle – Efficiency in Action | 电水壶案例 – 效率的实际计算

A student uses an electric kettle rated at 2000 W to heat 0.5 kg of water. The water temperature rises from 20°C to 100°C. The kettle is switched on for exactly 2 minutes (120 seconds). The specific heat capacity of water is 4200 J/(kg °C). Let’s calculate the energy efficiency of the kettle.

一位学生使用额定功率为 2000 W 的电水壶加热 0.5 kg 的水。水温从 20°C 上升到 100°C。水壶恰好接通 2 分钟（120 秒）。水的比热容为 4200 J/(kg °C)。我们来计算水壶的能量效率。

The electrical energy supplied to the kettle is calculated using E = P × t: E_input = 2000 W × 120 s = 240,000 J.

供给水壶的电能可以用 E = P × t 计算：E_输入 = 2000 W × 120 s = 240,000 J。

The useful thermal energy gained by the water is Q = m × c × Δθ: Δθ = 100°C – 20°C = 80°C. So Q = 0.5 kg × 4200 J/(kg °C) × 80°C = 168,000 J.

水获得的有用热能为 Q = m × c × Δθ：Δθ = 100°C – 20°C = 80°C。因此 Q = 0.5 kg × 4200 J/(kg °C) × 80°C = 168,000 J。

Therefore, the efficiency of the kettle is: Efficiency = (Q / E_input) × 100% = (168,000 / 240,000) × 100% = 70%.

因此，水壶的效率为：效率 = (Q / E_输入) ×

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