Heat, Light and Sound

Overview

Heat, Light and Sound form the core of classical physics taught at the upper-primary level and are essential for Bihar TET Paper II (Mathematics and Science). These topics test your understanding of energy transfer mechanisms, wave behaviour and everyday phenomena like echo, image formation and thermal expansion.

Questions typically appear in two forms: conceptual MCQs testing definitions and laws, and numerical problems involving mirror/lens formulae, heat calculations or wave equations. Mastering the fundamental principles here also strengthens your ability to answer pedagogy questions on science teaching methods—examiners often link content knowledge with how to demonstrate these concepts through practical activities.

The scope covers three distinct but related areas: modes of heat transfer (conduction, convection, radiation), behaviour of light (reflection, refraction, mirrors, lenses) and properties of sound (wave nature, speed, echo, characteristics). A solid grasp of formulae, ray diagrams and real-life applications is essential.

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Key Concepts

• **Heat is energy in transit** due to temperature difference; it flows from a hotter body to a colder body until thermal equilibrium is reached.

• **Three modes of heat transfer**: Conduction (through solids without particle movement), Convection (through fluids with particle movement), Radiation (through electromagnetic waves, no medium needed).

• **Light travels in straight lines** (rectilinear propagation); this explains shadows, eclipses and pinhole camera images.

• **Laws of Reflection**: (i) Angle of incidence = Angle of reflection. (ii) Incident ray, reflected ray and normal lie in the same plane.

• **Refraction** is the bending of light when it passes from one medium to another due to change in speed; governed by Snell's Law.

• **Sound is a mechanical longitudinal wave**; it requires a material medium (solid, liquid or gas) and cannot travel through vacuum.

• **Characteristics of sound**: Pitch (frequency), Loudness (amplitude), Quality/Timbre (waveform complexity).

• **Echo** is reflected sound heard after the original sound; minimum distance required is 17.2 m (at 20°C in air) for a distinct echo.

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Formulas / Key Facts

| Quantity / Law | Formula / Fact | |----------------|----------------| | Heat absorbed/released | Q = m × c × ΔT (m = mass, c = specific heat, ΔT = temperature change) | | Latent heat | Q = m × L (L = latent heat of fusion or vaporisation) | | Mirror formula | 1/f = 1/v + 1/u (f = focal length, v = image distance, u = object distance) | | Lens formula | 1/f = 1/v − 1/u (same sign convention as mirrors) | | Magnification (mirror/lens) | m = h'/h = −v/u (h' = image height, h = object height) | | Snell's Law | n = sin i / sin r (n = refractive index, i = angle of incidence, r = angle of refraction) | | Refractive index relation | n = c / v = Speed of light in vacuum / Speed of light in medium | | Speed of sound in air (20°C) | ≈ 343 m/s | | Wave equation | v = f × λ (v = velocity, f = frequency, λ = wavelength) | | Echo condition | Minimum distance = (Speed × Time) / 2; for distinct echo, time ≥ 0.1 s → d ≥ 17.2 m | | Audible frequency range | 20 Hz to 20,000 Hz (infrasound < 20 Hz; ultrasound > 20,000 Hz) |

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Worked Examples

### Example 1 — Heat Calculation **Problem**: How much heat is required to raise the temperature of 2 kg of water from 25°C to 75°C? (Specific heat of water = 4200 J/kg°C)

**Solution**:

• Given: m = 2 kg, c = 4200 J/kg°C, ΔT = 75 − 25 = 50°C
• Q = m × c × ΔT = 2 × 4200 × 50 = 4,20,000 J = 420 kJ

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### Example 2 — Mirror Formula **Problem**: An object is placed 30 cm in front of a concave mirror of focal length 15 cm. Find the image distance and magnification.

**Solution**:

• Sign convention (object on left): u = −30 cm, f = −15 cm
• 1/v = 1/f − 1/u = 1/(−15) − 1/(−30) = −1/15 + 1/30 = −2/30 + 1/30 = −1/30
• v = −30 cm (negative → image on same side as object → real image)
• Magnification m = −v/u = −(−30)/(−30) = −1 (inverted, same size)

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### Example 3 — Sound and Echo **Problem**: A person claps near a cliff and hears an echo after 0.4 s. If the speed of sound is 340 m/s, how far is the cliff?

**Solution**:

• Total distance travelled by sound = speed × time = 340 × 0.4 = 136 m
• Sound travels to cliff and back, so distance to cliff = 136 / 2 = 68 m

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Common Mistakes

| Wrong Thinking | Correct Fix | |----------------|-------------| | Confusing heat and temperature—treating them as the same. | Heat is energy transferred; temperature is a measure of average kinetic energy. A large block of ice has more heat capacity than a small cup of hot water even if its temperature is lower. | | Applying mirror formula without correct sign convention. | Always follow the New Cartesian Sign Convention: distances measured from pole; towards incident light is negative for object distance (u). Practice sign assignment before substituting. | | Believing sound travels fastest in air because we hear it in air. | Sound travels fastest in solids (particles closer), then liquids, slowest in gases. Steel ≈ 5000 m/s; Water ≈ 1500 m/s; Air ≈ 343 m/s. | | Forgetting to halve distance in echo problems. | Sound covers the distance twice (to reflector and back). Always divide total distance by 2. | | Mixing up frequency and amplitude effects on sound. | Frequency affects pitch (high/low); amplitude affects loudness (soft/loud). They are independent properties. |

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Quick Reference

• **Conduction**: Solids; **Convection**: Fluids; **Radiation**: No medium needed.
• Mirror/Lens formula: 1/f = 1/v ± 1/u — remember sign convention and whether + or − for lens.
• Real images are inverted; virtual images are erect.
• Speed of sound: Solid > Liquid > Gas.
• Audible range: 20 Hz – 20,000 Hz.
• Echo needs minimum 17.2 m distance (at 20°C in air).