Heat – Complete AMU 11 Notes – PYQ Junction Notes

        Heat is a scoring chapter in AMU — 1–2 questions every year. This expanded guide covers everything: temperature scales, thermometers, thermal expansion, anomalous expansion of water, calorimetry, latent heat, heat transfer, 15+ PYQs, 15+ numericals, and common mistakes. Let's master it! 🔥
    

1) Heat vs Temperature

Heat and Temperature are NOT the same thing — this is the #1 confusion for students. Heat is a form of energy (the total kinetic energy of all molecules), while temperature is a measure of the average kinetic energy of molecules.

Basis	Heat	Temperature
Definition	Form of energy that flows from hot to cold bodies	Measure of degree of hotness/coldness (avg KE of molecules)
Nature	Energy (total KE of all molecules)	Intensity (average KE per molecule)
SI Unit	Joule (J)	Kelvin (K)
Other units	calorie (1 cal = 4.184 J), kilocalorie	°Celsius, °Fahrenheit
Direction of flow	Flows from higher temperature to lower temperature	Determines direction of heat flow
Measuring device	Calorimeter	Thermometer
Effect of adding	Increases temperature OR changes state	Heat may increase or remain constant (during phase change)
Dependence	Depends on mass, specific heat, and temperature change	Does NOT depend on mass or type of substance

        🟡 Senior's Tip:

        Think of it like money in a bank:

        • Heat = total money in the account (depends on account size)

        • Temperature = how rich each person is (average per person)

        A large ocean has enormous HEAT but low TEMPERATURE. A small hot pan has high TEMPERATURE but less total HEAT.

🔥 AMU PYQ: SI unit of heat is:
(A) Kelvin (B) Celsius (C) Joule (D) Watt
Ans: C (heat is a form of energy → measured in Joule)

🔥 AMU PYQ: Temperature is a measure of:
(A) Total kinetic energy (B) Average kinetic energy (C) Potential energy (D) Heat content
Ans: B (temperature ∝ average KE of molecules)

2) Thermometers

A thermometer is a device used to measure temperature. It works on the principle that substances expand on heating and contract on cooling.

Types of Thermometers

Type	Working substance	Range	Use
Mercury thermometer	Mercury (silver liquid)	−10°C to 110°C (clinical) −30°C to 350°C (lab)	Clinical & laboratory use
Alcohol thermometer	Alcohol (coloured red/blue)	−80°C to 80°C	Very cold regions (mercury freezes at −39°C)
Digital thermometer	Thermistor / electronic sensor	−50°C to 300°C	Modern clinical & industrial

Construction & Working (Mercury Thermometer)

A fine, uniform capillary tube with a bulb at one end filled with mercury
When temperature rises → mercury expands → rises in capillary tube
When temperature falls → mercury contracts → falls in capillary tube
The tube is calibrated using fixed points: lower fixed point (ice point = 0°C) and upper fixed point (steam point = 100°C)
In clinical thermometer: a kink (constriction) prevents mercury from falling back — allows reading after removal

Fixed Points

Lower fixed point (Ice point): Temperature of pure melting ice at 1 atm pressure = 0°C / 32°F / 273 K
Upper fixed point (Steam point): Temperature of boiling water at 1 atm pressure = 100°C / 212°F / 373 K
The interval between them is divided into equal parts (e.g., 100 parts for Celsius)

        🟡 Senior's Tip:

        • Mercury is used because: it's opaque (easy to see), has uniform expansion, doesn't wet glass, high boiling point (357°C), low freezing point (−39°C)

        • Alcohol is used in cold regions because it freezes at −115°C (much lower than mercury's −39°C)

🔥 AMU PYQ: Which thermometer is used in very cold regions?
(A) Mercury (B) Alcohol (C) Digital (D) Both A and B
Ans: B (alcohol freezes at −115°C, mercury at −39°C)

3) Temperature Scales — Comparison & Conversion

There are three common temperature scales: Celsius, Fahrenheit, and Kelvin. Kelvin is the SI unit and is also called the absolute scale (no negative temperatures).

Property	Celsius (°C)	Fahrenheit (°F)	Kelvin (K)
Freezing point of water	0°C	32°F	273 K
Boiling point of water	100°C	212°F	373 K
Number of divisions	100	180	100
Absolute zero	−273.15°C	−459.67°F	0 K
SI unit?	No	No	Yes (SI base unit)
Used in	Most countries, science	USA, Myanmar, Liberia	Scientific research

Conversion formulas — MUST MEMORISE:
C/100 = (F − 32)/180 = (K − 273)/100
Celsius → Fahrenheit: °F = (°C × 9/5) + 32 Fahrenheit → Celsius: °C = (°F − 32) × 5/9 Celsius → Kelvin: K = °C + 273 Kelvin → Celsius: °C = K − 273 Fahrenheit → Kelvin: K = (F − 32) × 5/9 + 273

        🟡 Senior's Tip — Easy Tricks:

        • C → F: "Double and add 32" roughly: e.g., 25°C → 25×2 = 50, 50+30 = 80°F (actual: 77°F — close enough for estimates!)

        • F → C: "Subtract 32, then halve" roughly

        • K and °C: just add/subtract 273 — no complicated math!

Solved Example 1: Convert 77°F to Celsius.
C = (F − 32) × 5/9 = (77 − 32) × 5/9 = 45 × 5/9 = 25°C

Solved Example 2: Convert 30°C to Fahrenheit.
F = (C × 9/5) + 32 = (30 × 9/5) + 32 = 54 + 32 = 86°F

Solved Example 3: Convert 300 K to Celsius.
C = K − 273 = 300 − 273 = 27°C

Solved Example 4: At what temperature do Celsius and Fahrenheit scales coincide?
Let C = F = x x = (x − 32) × 5/9 9x = 5x − 160 4x = −160 x = −40 ∴ −40°C = −40°F (only point where both are equal!)

🔥 AMU PYQ: 100°C = ? °F
(A) 100°F (B) 180°F (C) 212°F (D) 273°F
Ans: C (100 × 9/5 + 32 = 212°F)

🔥 AMU PYQ: 0°C = ? K
(A) 0 K (B) 100 K (C) 273 K (D) 373 K
Ans: C (0 + 273 = 273 K)

🔥 AMU PYQ: Absolute zero in Celsius is:
(A) 0°C (B) −100°C (C) −273°C (D) −200°C
Ans: C (0 K = −273°C)

4) Thermal Expansion

When a substance is heated, its molecules vibrate more and move farther apart → the substance expands. Different materials expand by different amounts. Three types of expansion exist:

A. Linear Expansion (Length)

Change in length of a solid rod when heated. Applies to solids where one dimension is dominant (wires, rods, rails).

ΔL = α × L₀ × ΔT
ΔL = change in length, L₀ = original length, ΔT = temperature change α = coefficient of linear expansion (unit: K⁻¹ or °C⁻¹)
Final length L = L₀(1 + αΔT)

B. Superficial/Area Expansion (Surface)

Change in area of a solid surface when heated. Applies to sheets, plates, etc.

ΔA = β × A₀ × ΔT
β = coefficient of area (superficial) expansion β = 2α (relation between coefficients)
Final area A = A₀(1 + βΔT)

C. Cubical/Volumetric Expansion (Volume)

Change in volume of a substance when heated. Applies to solids, liquids, and gases.

ΔV = γ × V₀ × ΔT
γ = coefficient of volume (cubical) expansion γ = 3α (relation), γ = 3/2 β
Final volume V = V₀(1 + γΔT)

Type	Symbol	Formula	Applies to	Relation with α
Linear	α	ΔL = α L₀ ΔT	Solids (length)	α
Area / Superficial	β	ΔA = β A₀ ΔT	Solids (surface)	β = 2α
Volume / Cubical	γ	ΔV = γ V₀ ΔT	Solids, liquids, gases	γ = 3α

Applications of Thermal Expansion

Bimetallic strip: Two different metals (e.g., brass and invar/steel) bonded together. On heating, they expand by different amounts → strip bends. Used in: thermostats (automatic iron, AC, refrigerator), fire alarms, thermometers
Expansion gaps in railway tracks: Gaps left between rails to allow expansion in summer → prevents buckling
Expansion joints in bridges: One end of bridge is fixed, the other rests on rollers to allow expansion/contraction
Riveting: Hot rivets are used to join metal plates — on cooling, they contract and hold the plates tightly together
Iron rims on cart wheels: The rim is heated (expands), placed on the wheel, then cools (contracts) → tight fit
Loose cap on a bottle: Run hot water on the cap → cap expands → opens easily

        🟡 Senior's Tip — How to remember relations:

        • Linear (1D) → α

        • Area (2D) → β = 2α (think: 2 dimensions)

        • Volume (3D) → γ = 3α (think: 3 dimensions)

        • For liquids: we measure apparent expansion (in a container) vs real expansion (actual). Real expansion = apparent expansion + expansion of container

Solved Example: An iron rail is 10 m long at 20°C. α for iron = 1.2 × 10⁻⁵ K⁻¹. Find the increase in length on a hot day at 40°C.
ΔL = α L₀ ΔT = (1.2 × 10⁻⁵)(10)(40 − 20) = (1.2 × 10⁻⁵)(10)(20) = 1.2 × 10⁻⁵ × 200 = 240 × 10⁻⁵ = 0.0024 m = 2.4 mm

🔥 AMU PYQ: The coefficient of volumetric expansion γ is related to linear expansion α as:
(A) γ = α (B) γ = 2α (C) γ = 3α (D) γ = α/3
Ans: C (γ = 3α — volume has 3 dimensions)

🔥 AMU PYQ: Bimetallic strip is used in:
(A) Thermometer (B) Thermostat (C) Barometer (D) Manometer
Ans: B (thermostat — bimetallic strip bends to make/break electrical contact)

5) Anomalous Expansion of Water ⭐

Most substances expand when heated and contract when cooled — their density decreases with increasing temperature. Water is EXCEPTIONAL in the range 0°C to 4°C.

Key fact:
Water has its MAXIMUM DENSITY at 4°C
From 0°C → 4°C: water contracts (density INCREASES) From 4°C → above: water expands (density DECREASES) Ice (0°C) is LESS dense than water at 4°C → ICE FLOATS

Temperature	Behavior	Density
0°C (ice)	Solid — open crystal structure	0.917 g/cm³ (less — floats)
0°C → 4°C	Contracts (water shrinks)	Increases
4°C	Minimum volume	1.000 g/cm³ (maximum)
4°C → above	Expands normally	Decreases

Why is this important? — Lakes freeze from top!

In winter, air temperature drops → surface water cools → becomes denser → sinks (convection currents)
When water reaches 4°C, it is at maximum density → sinks to bottom
Surface water continues cooling below 4°C → becomes less dense → stays at top
At 0°C, surface water freezes into ice → ice is less dense → floats on top
The ice layer acts as insulator → water below stays at ~4°C → marine life survives

        🟡 Senior's Tip:

        • If water didn't have anomalous expansion, lakes would freeze from the BOTTOM → all aquatic life would die!

        • Ice floats on water because ice density (0.917 g/cm³) < water density (1.000 g/cm³)

        • This is why icebergs float — about 90% of an iceberg is underwater

🔥 AMU PYQ: Water has maximum density at:
(A) 0°C (B) 4°C (C) 100°C (D) −4°C
Ans: B (4°C — anomalous expansion)

🔥 AMU PYQ: Why do fish survive in frozen lakes?
(A) Ice is a good conductor (B) Ice layer insulates the water below (C) Fish hibernate (D) Water at bottom is at 0°C
Ans: B (ice insulates → water below remains at ~4°C)

6) Calorimetry & Specific Heat Capacity ⭐

Heat Capacity

The amount of heat required to raise the temperature of a body by 1°C (or 1 K).
Heat Capacity = m × c (unit: J/°C or J/K)

Specific Heat Capacity (c)

The amount of heat required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).
c = Q / (m × ΔT) Unit: J kg⁻¹ °C⁻¹ (or J/kg°C)

Fundamental formula — THE most important in this chapter:
Q = m × c × ΔT
Q = heat energy (J or cal) m = mass (kg or g) c = specific heat capacity (J/kg°C or cal/g°C) ΔT = change in temperature (°C or K)

Substance	Specific Heat (J/kg°C)	Specific Heat (cal/g°C)
Water	4186 ≈ 4200	1.0
Ice	2100	0.5
Aluminium	900	0.215
Iron/Steel	450	0.107
Copper	390	0.093
Silver	235	0.056
Sand	~800	~0.19
Alcohol	2400	0.58

        🟡 Senior's Tip — Why is water special?

        Water has the highest specific heat among common substances (4200 J/kg°C). This means:

        • Water heats up slowly (takes lots of energy)

        • Water cools down slowly (releases lots of energy)

        • This is why coastal areas have moderate climate — sea water absorbs heat in summer, releases in winter

        • This is why water is used as coolant in car radiators and in thermal power plants

Principle of Calorimetry

Heat lost by hot body = Heat gained by cold body m₁ × c₁ × (T_hot − T_final) = m₂ × c₂ × (T_final − T_cold)
m₁, c₁ = mass and specific heat of hot body m₂, c₂ = mass and specific heat of cold body T_final = equilibrium temperature Assumption: No heat is lost to the surroundings (ideal calorimeter)

Solved Example 1 (Calorimetry):
2 kg of water at 80°C is mixed with 3 kg of water at 20°C. Find final temperature. (c_water = 4200 J/kg°C)
Heat lost = Heat gained 2 × 4200 × (80 − T) = 3 × 4200 × (T − 20) 2(80 − T) = 3(T − 20) 160 − 2T = 3T − 60 160 + 60 = 5T T = 220/5 = 44°C

Solved Example 2 (Calorimetry — different substances):
A 0.5 kg iron block at 200°C is placed in 1 kg water at 25°C. Find final temperature.
c_iron = 450 J/kg°C, c_water = 4200 J/kg°C
Heat lost by iron = Heat gained by water 0.5 × 450 × (200 − T) = 1 × 4200 × (T − 25) 225(200 − T) = 4200(T − 25) 45000 − 225T = 4200T − 105000 45000 + 105000 = 4200T + 225T 150000 = 4425T T ≈ 33.9°C

🔥 AMU PYQ: Which has higher specific heat — water or sand?
(A) Water (B) Sand (C) Same (D) None
Ans: A (water = 4200 J/kg°C, sand ≈ 800 J/kg°C. Sea stays cool during day, warm at night)

🔥 AMU PYQ: SI unit of specific heat capacity is:
(A) J/kg (B) J/°C (C) J/kg°C (D) J/g°C
Ans: C (Joule per kilogram per degree Celsius)

🔥 AMU PYQ: Heat required to raise temperature of 2 kg water from 30°C to 50°C is:
(A) 168 kJ (B) 168 J (C) 4200 J (D) 84 kJ
Ans: A (Q = 2 × 4200 × 20 = 168,000 J = 168 kJ)

7) Latent Heat

Latent heat is the heat absorbed or released when a substance changes its state (solid ⇌ liquid ⇌ gas) without changing its temperature.

Q = m × L
Q = heat energy (J or cal) m = mass (kg or g) L = latent heat (J/kg or cal/g)

Type	Symbol	Process	Definition	Value (water)
Latent heat of Fusion	L_f	Solid ↔ Liquid	Heat to melt 1 kg of ice at 0°C → water at 0°C (or freeze reverse)	3.36 × 10⁵ J/kg = 336 kJ/kg = 80 cal/g
Latent heat of Vaporization	L_v	Liquid ↔ Gas	Heat to boil 1 kg of water at 100°C → steam at 100°C (or condense reverse)	22.6 × 10⁵ J/kg = 2260 kJ/kg = 540 cal/g

        🟡 Senior's Tip — Key Insight:

        During phase change, temperature does NOT change. All the heat energy goes into breaking the intermolecular bonds (to separate molecules) rather than increasing kinetic energy.

        • Melting: heat breaks the crystal lattice bonds

        • Boiling: heat overcomes the intermolecular attractive forces completely

        • Lv > Lf because gas molecules are completely separated (much more energy needed)

Solved Example: How much heat is needed to melt 2 kg of ice at 0°C?
Q = m × L_f = 2 × 3.36 × 10⁵ = 6.72 × 10⁵ J = 672 kJ

Solved Example: How much heat is needed to convert 0.5 kg of water at 100°C to steam at 100°C?
Q = m × L_v = 0.5 × 22.6 × 10⁵ = 11.3 × 10⁵ J = 1.13 × 10⁶ J

🔥 AMU PYQ: During change of state, temperature:
(A) Increases (B) Decreases (C) Remains constant (D) First increases then decreases
Ans: C (heat is used to break bonds, not raise temperature — latent heat)

🔥 AMU PYQ: Latent heat of fusion of ice is:
(A) 336 J/kg (B) 3360 J/kg (C) 3.36 × 10⁵ J/kg (D) 22.6 × 10⁵ J/kg
Ans: C (L_f = 3.36 × 10⁵ J/kg = 336 kJ/kg)

🔥 AMU PYQ: Ice at 0°C is better coolant than water at 0°C because:
(A) Ice is colder (B) Ice absorbs latent heat to melt (C) Ice is denser (D) None
Ans: B (ice absorbs L_f = 336 kJ/kg while melting, providing extra cooling)

8) Evaporation vs Boiling

Both evaporation and boiling involve conversion of liquid to vapour, but they are DIFFERENT processes.

Basis	Evaporation	Boiling
Definition	Surface phenomenon — liquid changes to vapour at the surface	Bulk phenomenon — liquid changes to vapour throughout the liquid
Temperature	Occurs at any temperature below boiling point	Occurs at a fixed temperature (boiling point)
Where it happens	Only at the surface of the liquid	Throughout the entire volume of the liquid (bubbles form)
Speed	Slow process	Fast process
Effect of pressure	Increases with wind speed, surface area, temperature	Boiling point increases with external pressure
Cooling effect	Causes cooling (molecules with higher KE escape → avg KE decreases)	No significant cooling of remaining liquid
Examples	Drying of clothes, sweat evaporating from skin, water cycle	Water boiling for tea, cooking rice

Factors Affecting Rate of Evaporation

Temperature: Higher temperature → faster evaporation (more molecules have enough KE to escape)
Surface area: Larger surface area → faster evaporation (more molecules at surface)
Wind speed: Higher wind → faster evaporation (removes vapour from surface → maintains concentration gradient)
Humidity: Lower humidity → faster evaporation (air can hold more water vapour)
Nature of liquid: More volatile liquids (alcohol, petrol) evaporate faster than less volatile ones (water, oil)

        🟡 Senior's Tip — Cooling by evaporation:

        • When you sweat, water evaporates from your skin → takes away latent heat of vaporization from your body → you feel cool

        • Desert coolers (evaporative coolers) work on this principle — water evaporates, cooling the air

        • Interesting fact: Alcohol evaporates faster than water → it feels colder when rubbed on skin because it takes away heat faster!

🔥 AMU PYQ: Evaporation causes cooling because:
(A) Heat is absorbed from surroundings (B) Heat is released (C) Temperature increases (D) Vapour is hot
Ans: A (molecules with higher KE escape → remaining molecules have lower avg KE → cooling)

9) Heat Transfer — Conduction, Convection & Radiation ⭐

Heat can be transferred in three ways — conduction, convection, and radiation. Understanding their differences is crucial for AMU.

Property	Conduction	Convection	Radiation
Medium required	Yes (solids best)	Yes (liquids & gases)	No (vacuum also works)
Mechanism	Molecular vibrations + electron flow — molecules vibrate and pass energy to neighbours. No bulk movement.	Physical movement of molecules — hot fluid rises (expands → less dense), cool fluid sinks → convection currents	Electromagnetic waves (infrared radiation) — travels at speed of light
Speed	Slow	Moderate	Fastest (speed of light: 3×10⁸ m/s)
Example	Iron rod getting hot at one end when other end is in fire Hand on a hot pan	Boiling water (convection currents) Sea breeze, land breeze AC placement (above → cool air sinks)	Sun's heat reaching Earth Campfire warmth Heat from a room heater (infrared)
Depends on	Thermal conductivity (k), area, temp difference, thickness	Density differences, gravity, fluid properties	Temperature, surface area, nature of surface (colour, texture)

A. Conduction — Detailed

Thermal conductivity (k): A measure of how well a material conducts heat. Units: W/mK or W/m°C.

Rate of heat conduction (Fourier's law):
Q/t = k × A × (T₁ − T₂) / d
Q/t = rate of heat transfer (W or J/s) k = thermal conductivity of material A = cross-sectional area, d = thickness T₁ − T₂ = temperature difference

Good Conductors	k (W/mK)	Bad Conductors (Insulators)	k (W/mK)
Silver	430	Wood	0.1–0.2
Copper	390	Rubber	0.15
Gold	310	Glass wool	0.04
Aluminium	210	Air	0.025
Iron	80	Plastic	0.2

        🟡 Senior's Tip — Applications:

        • Good conductors → used where heat needs to flow quickly: Cooking utensils (Cu/Al bottom), engine radiators (Cu), heat sinks in electronics (Al)

        • Bad conductors → used where heat needs to be trapped/prevented: Cooking utensil handles (wood/plastic), ice box walls (glass wool), winter clothes (wool — traps air), house insulation (glass wool, foam)

        • Why do woollen clothes keep us warm? Wool traps a layer of air. Air is a bad conductor → body heat can't escape → we stay warm. (Wool does NOT give heat!)

Solved Example: A copper rod of area 0.01 m² and length 0.5 m has ends at 100°C and 0°C. k_Cu = 390 W/mK. Find rate of heat flow.
Q/t = k × A × ΔT / d = 390 × 0.01 × 100 / 0.5 = 390 × 0.01 × 200 = 780 W (Joules per second)

B. Convection — Detailed

Convection occurs in fluids (liquids and gases). It involves actual movement of molecules.

Type	Mechanism	Examples
Natural (Free) Convection	Fluid moves due to density differences caused by temperature gradients — gravity driven. No external force.	Sea breeze, land breeze Hot air rising from a radiator Boiling water (water at bottom gets hot → rises → cools → sinks)
Forced Convection	Fluid is forced to move by an external agent (fan, pump, blower)	Ceiling fan (circulates air) Car radiator with fan CPU cooling fan Air conditioner blower

Sea Breeze & Land Breeze — Complete Explanation

Day (Sea Breeze):
- Land heats up faster than sea water (land has low specific heat, water has high specific heat)
- Air above land becomes hot → expands → rises (density decreases)
- Cool air from above the sea rushes in towards the land to replace the rising hot air
- This cool wind from sea to land is called sea breeze
Night (Land Breeze):
- Land cools down faster than sea water at night
- Air above the sea is now warmer (water cools slowly) → rises
- Cool air from the land rushes towards the sea
- This cool wind from land to sea is called land breeze

        🟡 Senior's Tip — AC vs Heater Placement:

        • Air conditioner (AC) is placed high on the wall — cool air is denser → sinks → cools the room from top to bottom

        • Room heater is placed near the floor — hot air is less dense → rises → heats the room from bottom to top

        • 🔥 This is a VERY common AMU MCQ!

C. Radiation — Detailed

Radiation does NOT require any medium — it can travel through vacuum
It travels as electromagnetic waves (primarily infrared, also visible and UV for very hot objects)
All objects emit radiation — hotter objects emit more radiation and at shorter wavelengths
A black body is a perfect absorber and perfect emitter of radiation

Kirchhoff's Law (of thermal radiation):
Good absorbers of radiation are also good emitters
A black surface that absorbs all radiation also emits radiation efficiently A polished/shiny surface that reflects radiation is a poor emitter

Application	Explanation
Black car gets hotter in summer	Black absorbs more radiation (good absorber) → gets hotter than a white car
White/light coloured clothes in summer	White reflects radiation (poor absorber) → keeps you cool
Dark coloured clothes in winter	Dark absorbs radiation (good absorber) → keeps you warm
Solar water heater has black painted surface	Black absorbs maximum solar radiation → heats water efficiently
Cooking utensils have polished/ shiny outer surface	Polished surface is poor radiator → keeps heat inside the utensil
Greenhouse effect	Glass lets sunlight in (visible light) but traps infrared radiation → inside gets hot

Key comparison — Good Radiator vs Good Reflector:
Dull, rough, dark surfaces → good absorbers AND good emitters (radiators) Shiny, smooth, light surfaces → poor absorbers AND poor emitters (good reflectors)

🔥 AMU PYQ: The best conductor of heat is:
(A) Iron (B) Silver (C) Copper (D) Aluminium
Ans: B (silver has highest thermal conductivity: 430 W/mK)

🔥 AMU PYQ: Heat from the Sun reaches Earth by:
(A) Conduction (B) Convection (C) Radiation (D) Both A & B
Ans: C (radiation — no medium needed for vacuum)

🔥 AMU PYQ: A room heater is placed near the floor because:
(A) Hot air rises (B) Hot air sinks (C) Radiation works better (D) Conduction is faster
Ans: A (hot air is less dense → rises → heats the room by convection)

🔥 AMU PYQ: In a desert cooler, the cooling is due to:
(A) Conduction (B) Convection (C) Evaporation (D) Radiation
Ans: C (evaporation of water absorbs latent heat → cools the air)

10) PYQ Battle — 18 AMU Past Questions on Heat

Q1. SI unit of heat is:
(A) Kelvin (B) Celsius (C) Joule (D) Watt
Ans: C

Q2. 100°C = ? °F
(A) 100°F (B) 180°F (C) 212°F (D) 273°F
Ans: C (100 × 9/5 + 32 = 212)

Q3. 0°C = ? K
(A) 0 K (B) 100 K (C) 273 K (D) 373 K
Ans: C

Q4. Absolute zero in Celsius is:
(A) 0°C (B) −100°C (C) −273°C (D) −200°C
Ans: C

Q5. Which has higher specific heat — water or sand?
(A) Water (B) Sand (C) Same (D) None
Ans: A (water = 4200, sand ≈ 800 J/kg°C)

Q6. SI unit of specific heat capacity is:
(A) J/kg (B) J/°C (C) J/kg°C (D) J/g°C
Ans: C

Q7. During change of state, temperature:
(A) Increases (B) Decreases (C) Remains constant (D) First increases then decreases
Ans: C (latent heat)

Q8. Latent heat of fusion of ice is:
(A) 336 J/kg (B) 3360 J/kg (C) 3.36 × 10⁵ J/kg (D) 22.6 × 10⁵ J/kg
Ans: C

Q9. Water has maximum density at:
(A) 0°C (B) 4°C (C) 100°C (D) −4°C
Ans: B (anomalous expansion of water)

Q10. The best conductor of heat is:
(A) Iron (B) Silver (C) Copper (D) Aluminium
Ans: B

Q11. Heat from Sun reaches Earth by:
(A) Conduction (B) Convection (C) Radiation (D) Both A & B
Ans: C

Q12. A room heater is placed near the floor because:
(A) Hot air rises (B) Hot air sinks (C) Radiation is better (D) Conduction is faster
Ans: A

Q13. The coefficient of volumetric expansion γ = ?
(A) α (B) 2α (C) 3α (D) α/3
Ans: C

Q14. Bimetallic strip is used in:
(A) Thermometer (B) Thermostat (C) Barometer (D) Manometer
Ans: B

Q15. Evaporation causes cooling because:
(A) Heat is absorbed (B) Heat is released (C) Temp increases (D) Vapour is hot
Ans: A

Q16. Why do fish survive in frozen lakes?
(A) Ice conducts heat (B) Ice insulates (C) Fish hibernate (D) Bottom water is 0°C
Ans: B (ice layer insulates → water below at ~4°C)

Q17. Convert −40°C to Fahrenheit.
Ans: F = (−40 × 9/5) + 32 = −72 + 32 = −40°F (coincident point!)

Q18. Heat required to raise 2 kg water from 30°C to 50°C is:
Ans: Q = 2 × 4200 × 20 = 168,000 J = 168 kJ

Q19. Which of the following is NOT a mode of heat transfer?
(A) Conduction (B) Convection (C) Radiation (D) Refraction
Ans: D (refraction is for light, not heat transfer)

Q20. A thermos flask (vacuum flask) minimizes heat loss by:
(A) All three modes (B) Only conduction (C) Only radiation (D) Only convection
Ans: A — vacuum stops conduction & convection, silvered walls reflect radiation

Q21. Which of these has the highest specific heat capacity?
(A) Iron (B) Copper (C) Water (D) Aluminium
Ans: C (water = 4200 J/kg°C, much higher than metals)

Q22. The process in which heat is transferred by electromagnetic waves is:
(A) Conduction (B) Convection (C) Radiation (D) Evaporation
Ans: C

Q23. Two rods of same material and length have area ratio 1:2. Find ratio of heat conducted (same ΔT).
Ans: Q/t ∝ A → ratio = 1:2

11) Heating Curve of Ice — Complete State Change Graph ⭐

The heating curve of ice is a graph of temperature vs time when a block of ice at sub-zero temperature is heated continuously until it becomes steam. This is a FAVOURITE AMU topic!

Stage-wise breakdown (starting from −20°C ice to 100°C+ steam):
Stage A→B: Ice at −20°C → 0°C (Q = m c_ice ΔT, no state change) Stage B→C: Ice at 0°C melts → water at 0°C (Q = m L_f, temp constant) Stage C→D: Water at 0°C → 100°C (Q = m c_water ΔT, no state change) Stage D→E: Water at 100°C boils → steam at 100°C (Q = m L_v, temp constant) Stage E→F: Steam at 100°C → above (Q = m c_steam ΔT, if continued)

Stage	Change	Temp	Formula	Heat type
A→B	Ice warming	−20°C → 0°C	Q = mcΔT (c_ice = 2100)	Sensible
B→C	Ice → Water (melting)	0°C constant	Q = mL_f	Latent (fusion)
C→D	Water warming	0°C → 100°C	Q = mcΔT (c_water = 4200)	Sensible
D→E	Water → Steam (boiling)	100°C constant	Q = mL_v	Latent (vaporization)
E→F	Steam heating	100°C → higher	Q = mcΔT (c_steam ≈ 2000)	Sensible

        🟡 Senior's Tip — Why is the graph flat at B→C and D→E?

        During melting (B→C) and boiling (D→E), temperature DOES NOT RISE. All the heat supplied is used to break intermolecular bonds (latent heat). No increase in kinetic energy → no temperature increase. The flat regions in the graph are called plateaus or temperature arrest points.

🔥 AMU PYQ: In the heating curve of ice, temperature remains constant during:
(A) A→B only (B) B→C and D→E (C) C→D only (D) E→F only
Ans: B (melting and boiling — latent heat stages)

12) Practice Problems — 23 Numericals with Step-by-Step Solutions

Problem 1: Convert 104°F to Celsius.
C = (104 − 32) × 5/9 = 72 × 5/9 = 40°C

Problem 2: Convert 50°C to Kelvin.
K = 50 + 273 = 323 K

Problem 3: Convert 400 K to Fahrenheit.
C = 400 − 273 = 127°C F = (127 × 9/5) + 32 = 228.6 + 32 = 260.6°F

Problem 4: A 5 kg iron block (c = 450 J/kg°C) is heated from 25°C to 75°C. Find heat required.
Q = m c ΔT = 5 × 450 × (75 − 25) = 5 × 450 × 50 = 112,500 J

Problem 5: 300 g of water at 60°C is mixed with 200 g of water at 20°C. Find final temperature.
Let T = final temp 300 × (60 − T) = 200 × (T − 20) 18000 − 300T = 200T − 4000 22000 = 500T T = 44°C

Problem 6: 2 kg of water at 90°C is mixed with 3 kg of water at 10°C. Find T.
2(90 − T) = 3(T − 10) 180 − 2T = 3T − 30 210 = 5T T = 42°C

Problem 7: A 0.2 kg copper block at 150°C is dropped into 0.5 kg water at 25°C. c_Cu = 390 J/kg°C, c_w = 4200 J/kg°C. Find T.
0.2 × 390 × (150 − T) = 0.5 × 4200 × (T − 25) 78(150 − T) = 2100(T − 25) 11700 − 78T = 2100T − 52500 64200 = 2178T T ≈ 29.5°C

Problem 8: Heat required to melt 500 g of ice at 0°C. (L_f = 336 kJ/kg)
Q = m × L_f = 0.5 × 336 × 10³ = 168,000 J = 168 kJ

Problem 9: Heat required to convert 100 g of water at 100°C to steam. (L_v = 2260 kJ/kg)
Q = 0.1 × 2260 × 10³ = 226,000 J

Problem 10: Total heat to convert 200 g of ice at −10°C to steam at 100°C.
Step 1: Heat ice from −10°C to 0°C: Q₁ = 0.2 × 2100 × 10 = 4200 J Step 2: Melt ice at 0°C: Q₂ = 0.2 × 336 × 10³ = 67,200 J Step 3: Heat water from 0°C to 100°C: Q₃ = 0.2 × 4200 × 100 = 84,000 J Step 4: Boil water at 100°C: Q₄ = 0.2 × 2260 × 10³ = 452,000 J Total = 4200 + 67,200 + 84,000 + 452,000 = 607,400 J

Problem 11: An iron rod (α = 1.2 × 10⁻⁵ K⁻¹) is 5 m long at 20°C. Find length at 60°C.
ΔL = α L₀ ΔT = 1.2 × 10⁻⁵ × 5 × 40 = 2.4 × 10⁻³ m = 2.4 mm L = 5 + 0.0024 = 5.0024 m

Problem 12: A copper wire (α = 1.7 × 10⁻⁵ K⁻¹) is 2 m long at 30°C. At what temperature will it be 2.0034 m?
ΔL = 2.0034 − 2 = 0.0034 m ΔT = ΔL / (α L₀) = 0.0034 / (1.7×10⁻⁵ × 2) = 0.0034 / 3.4×10⁻⁵ = 100°C T = 30 + 100 = 130°C

Problem 13: Calculate the rate of heat conduction through a glass window (area = 2 m², thickness = 4 mm, k = 0.8 W/mK) when inside temp = 25°C, outside = 5°C.
Q/t = k A ΔT / d = 0.8 × 2 × 20 / 0.004 = 8000 W

Problem 14: If 1 kg of water at 30°C is mixed with 1 kg of ice at 0°C, what is the final temperature? Assume all ice melts first if possible.
Heat available from water cooling to 0°C: Q_w = 1 × 4200 × 30 = 126,000 J Heat needed to melt ice: Q_ice = 1 × 336,000 = 336,000 J Since Q_w < Q_ice, not all ice melts. Final temp = 0°C (mixture of ice + water) Mass of ice melted = Q_w / L_f = 126,000 / 336,000 = 0.375 kg

Problem 15: A copper rod (k = 390 W/mK) of area 0.005 m² and length 0.4 m conducts heat from 100°C to 0°C. Find the rate of heat flow.
Q/t = 390 × 0.005 × 100 / 0.4 = 390 × 0.005 × 250 = 487.5 W

Problem 16: Convert 98.6°F (human body temperature) to Celsius.
C = (98.6 − 32) × 5/9 = 66.6 × 5/9 = 37°C

Problem 17: A black car absorbs 80% of incident solar radiation. If incident power is 1000 W/m² and car area is 6 m², how much heat is absorbed per second?
Heat absorbed = 0.8 × 1000 × 6 = 4800 W = 4800 J/s

Problem 18: 4 kg of water at 80°C is mixed with 2 kg of water at 20°C. Find the final temperature and the heat exchanged.
4(80 − T) = 2(T − 20) 320 − 4T = 2T − 40 360 = 6T T = 60°C Heat exchanged = 4 × 4200 × (80 − 60) = 4 × 4200 × 20 = 336,000 J

Problem 19: A steel bridge is 100 m long at 15°C. On a hot summer day at 45°C, how much does it expand? (α_steel = 1.2 × 10⁻⁵ K⁻¹)
ΔL = α L₀ ΔT = 1.2 × 10⁻⁵ × 100 × 30 = 1.2 × 10⁻⁵ × 3000 = 0.036 m = 3.6 cm

Problem 20: A glass vessel of volume 500 ml is filled with mercury at 20°C. If temperature rises to 60°C, how much mercury overflows? (γ_mercury = 1.8 × 10⁻⁴ K⁻¹, γ_glass = 2.5 × 10⁻⁵ K⁻¹)
Apparent expansion of mercury = γ_mercury − γ_glass = 1.8 × 10⁻⁴ − 0.25 × 10⁻⁴ = 1.55 × 10⁻⁴ K⁻¹ Overflow = V₀ × γ_app × ΔT = 500 × 1.55 × 10⁻⁴ × 40 = 500 × 6.2 × 10⁻³ = 3.1 ml

Problem 21: How much ice at 0°C must be added to 0.5 kg of water at 30°C to cool it to 0°C? (L_f = 336 kJ/kg)
Heat lost by water = m_w × c_w × ΔT = 0.5 × 4200 × 30 = 63,000 J Heat gained by ice to melt = m_ice × L_f m_ice = 63,000 / 336,000 = 0.1875 kg = 187.5 g

Problem 22: A geyser heats 2 kg of water from 20°C to 80°C in 5 minutes. Find the power rating. (c_w = 4200 J/kg°C)
Q = m c ΔT = 2 × 4200 × 60 = 504,000 J Power = Q / t = 504,000 / (5 × 60) = 504,000 / 300 = 1680 W

Problem 23: A room has a window of area 3 m² and thickness 5 mm. Inside temp = 25°C, outside = 5°C. k_glass = 0.8 W/mK. Find heat lost per hour.
Q/t = k A ΔT / d = 0.8 × 3 × 20 / 0.005 = 9600 W = 9600 J/s Q per hour = 9600 × 3600 = 34.56 × 10⁶ J = 34.56 MJ

13) Greenhouse Effect — Detailed

The greenhouse effect is the process by which the Earth's atmosphere traps heat from the Sun, keeping the planet warm enough for life to exist.

Mechanism:

Sunlight (visible light + UV) passes through the atmosphere and reaches Earth's surface
The Earth absorbs this radiation and re-emits it as infrared radiation (heat)
Greenhouse gases (CO₂, CH₄, H₂O vapour) in the atmosphere absorb infrared radiation and re-radiate it back towards Earth
This traps heat in the atmosphere → Earth stays warm (avg ~15°C instead of −18°C without greenhouse effect)

Greenhouse Gas	Source	Contribution
CO₂	Burning fossil fuels, deforestation	~60%
CH₄ (Methane)	Agriculture, livestock, landfills	~20%
H₂O vapour	Natural (evaporation)	~10%
CFCs	Refrigerants, aerosols	~10%

        🟡 Senior's Tip — Differentiate:

        • Natural greenhouse effect — essential for life (keeps Earth warm)

        • Enhanced greenhouse effect — human activities increase greenhouse gases → global warming (melting glaciers, rising sea levels, climate change)

        • Glass in a greenhouse/ car: Glass lets visible light in, but traps infrared → inside gets hot (same principle as Earth's atmosphere)

🔥 AMU PYQ: The greenhouse effect is caused primarily by:
(A) CO₂ (B) O₂ (C) N₂ (D) H₂
Ans: A

14) Thermal Expansion in Liquids — Apparent vs Real Expansion

When a liquid is heated in a container, both the liquid AND the container expand. What we observe is the apparent expansion of the liquid (the net effect).

γ_real = γ_apparent + γ_container
γ_real = actual expansion coefficient of liquid γ_apparent = what we observe (rise in container) γ_container = expansion coefficient of the container material (≈ 3α_solid)

        🟡 Senior's Tip:

        • For liquids: γ_real is always greater than γ_apparent because the container also expands

        • Water is ANOMALOUS in expansion behavior (explained in section 5)

        • For gases: all gases have nearly the same γ ≈ 1/273 K⁻¹ (Charles' law)

15) Common Mistakes ❌ — Don't Lose Marks!

❌ Mistake 1: Confusing heat and temperature.
✅ Fix: Heat is energy (J). Temperature is measure of avg KE (K). Different concepts!

❌ Mistake 2: Forgetting to convert °C to K in scientific formulas.
✅ Fix: For ΔT (temperature DIFFERENCE), °C and K are same magnitude. But absolute temperatures must be in K for gas laws. For specific heat: ΔT in °C is fine since 1°C = 1 K.

❌ Mistake 3: Using wrong units (g instead of kg).
✅ Fix: Always convert grams to kg: mass in kg, c in J/kg°C, Q in J. If using g: c in J/g°C, Q in J. Stay consistent!

❌ Mistake 4: Thinking temperature changes during phase change.
✅ Fix: Temperature is CONSTANT during melting/boiling. All heat is used as latent heat (breaking bonds).

❌ Mistake 5: Confusing conduction vs convection vs radiation.
✅ Fix: Conduction = solids (no movement of material). Convection = fluids (movement of material). Radiation = vacuum (EM waves).

❌ Mistake 6: Forgetting that radiation doesn't need a medium.
✅ Fix: Sun's heat reaches Earth via radiation through vacuum. Only mode that works without medium.

❌ Mistake 7: Thinking good conductors and good radiators are the same.
✅ Fix: Good conductors (metals) ≠ good radiators. Dull black surfaces are good radiators, shiny surfaces are poor radiators.

❌ Mistake 8: Mixing up α, β, γ coefficients.
✅ Fix: β = 2α (2D), γ = 3α (3D). Easy: linear → 1 coefficient, area → 2×, volume → 3×.

❌ Mistake 9: Assuming all substances expand the same way as water.
✅ Fix: Most substances expand when heated, contract when cooled. Water is ANOMALOUS between 0°C–4°C (contracts when heated).

❌ Mistake 10: Forgetting to include all steps in mixed problems (ice → water → steam).
✅ Fix: When ice goes to steam: heat ice to 0°C + melt ice + heat water to 100°C + boil water. Each step has its own formula!

16) Quick Revision — One-Liners

Heat = energy (J). Temperature = avg KE of molecules (K, °C, °F)
F = C × 9/5 + 32 and K = C + 273 — memorize cold
C/100 = (F−32)/180 = (K−273)/100 — the universal conversion formula
−40°C = −40°F — the only temperature where both scales agree
Absolute zero = 0 K = −273.15°C — lowest possible temperature
Q = mcΔT — heat when temperature changes (specific heat)
Q = mL — heat during phase change (latent heat)
Water c = 4200 J/kg°C — highest among common substances
L_f (ice) = 336 kJ/kg — heat to melt ice at 0°C
L_v (water) = 2260 kJ/kg — heat to boil water at 100°C
Calorimetry: Heat lost = Heat gained (m₁c₁ΔT₁ = m₂c₂ΔT₂)
Thermal expansion: ΔL = αL₀ΔT, β = 2α, γ = 3α
Anomalous expansion of water: Max density at 4°C. Ice floats → lakes freeze from top
Conduction: solids, molecular vibration. Convection: fluids, mass movement. Radiation: EM waves, no medium
Good conductors: Silver > Copper > Gold > Aluminium > Iron
Insulators: Wood, plastic, rubber, air, glass wool, wool
Sea breeze (day): cool air from sea → land. Land breeze (night): cool air from land → sea
AC high (sinks cool air): Heater low (rises hot air) — convection
Black body: Perfect absorber = perfect emitter (Kirchhoff's law)
Evaporation: Cools (surface phenomenon, any temp). Boiling: Bulk phenomenon, fixed temp
Evaporation factors: Temp↑, Area↑, Wind↑, Humidity↓ → all increase evaporation

        ✅ Heat — Fully Covered! You now have a complete reference with theory, applications, 18 PYQs, 18 numericals, and 10 common mistakes. Focus on: Q = mcΔT, Q = mL, calorimetry, temperature conversion, and heat transfer modes. Solve the practice problems and you're exam-ready for AMU! 🔥📚
    