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Forces and Energy - Forms of Energy (Light, Sound, and Heat)

Grade 4IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Light energy travels in straight lines as waves. It can be reflected (bouncing off), refracted (bending when passing through different media), or absorbed. The speed of light is denoted by c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}.

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Sound energy is produced by vibrations and moves as longitudinal waves. It requires a medium (solid, liquid, or gas) to travel. Pitch is determined by the frequency (ff) of the vibration, while volume is determined by the amplitude (AA).

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Heat energy, or thermal energy, is the flow of energy from an object at a higher temperature (T1T_1) to an object at a lower temperature (T2T_2). This transfer continues until thermal equilibrium is reached where T1=T2T_1 = T_2.

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Materials can be classified by how they interact with light: transparent (all light passes), translucent (some light passes), and opaque (no light passes, creating a shadow where light is blocked).

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Heat transfer occurs through three main methods: conduction (solids), convection (liquids and gases), and radiation (electromagnetic waves).

📐Formulae

v=f×λv = f \times \lambda

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

Speed=Distance(d)Time(t)\text{Speed} = \frac{\text{Distance} (d)}{\text{Time} (t)}

💡Examples

Problem 1:

If a sound wave has a frequency of f=250 Hzf = 250 \text{ Hz} and a wavelength of λ=1.4 m\lambda = 1.4 \text{ m}, calculate the speed (vv) of the sound wave.

Solution:

v=250×1.4=350 m/sv = 250 \times 1.4 = 350 \text{ m/s}

Explanation:

By using the wave equation v=fλv = f \lambda, we multiply the frequency by the wavelength to find that the sound is traveling at 350 m/s350 \text{ m/s} through the air.

Problem 2:

A cup of cocoa is at 60∘C60^\circ C and the surrounding air is 22∘C22^\circ C. In which direction will the heat energy flow?

Solution:

From the cocoa (60∘C60^\circ C) to the air (22∘C22^\circ C).

Explanation:

Heat energy always moves from a higher temperature (ThighT_{high}) to a lower temperature (TlowT_{low}) until they reach the same temperature.

Problem 3:

Calculate the temperature in Kelvin (KK) if a thermometer reads 25∘C25^\circ C.

Solution:

T(K)=25+273.15=298.15 KT(K) = 25 + 273.15 = 298.15 \text{ K}

Explanation:

To convert Celsius to Kelvin, we use the formula T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15.

Forms of Energy (Light, Sound, and Heat) Grade 4 Notes & Examples