What is the difference between heat and temperature?

Heat is the energy transferred between objects due to temperature difference, measured in joules. Temperature is a measure of average kinetic energy of molecules, measured in Kelvin, Celsius, or Fahrenheit. Heat flows from higher temperature to lower temperature.

What is thermal expansion and its types?

Thermal expansion is the tendency of matter to increase in volume when heated. There are three types: Linear expansion (ΔL = αL₀ΔT), Area expansion (ΔA = βA₀ΔT where β = 2α), and Volume expansion (ΔV = γV₀ΔT where γ = 3α).

What is specific heat capacity?

Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 Kelvin. Formula: Q = mcΔT, where c is specific heat capacity. Water has highest specific heat capacity (4200 J/kg·K).

Latent heat is the heat absorbed or released during phase change at constant temperature. Latent heat of fusion (melting): Q = mL_f. Latent heat of vaporization (boiling): Q = mL_v. For water, L_f = 334 kJ/kg and L_v = 2260 kJ/kg.

What are the three modes of heat transfer?

The three modes are: 1) Conduction - heat transfer through direct contact (Q/t = kA(T₁-T₂)/L), 2) Convection - heat transfer through fluid movement, 3) Radiation - heat transfer through electromagnetic waves (Stefan's law: P = σAT⁴).

What is Newton's law of cooling?

Newton's law of cooling states that the rate of cooling of a body is proportional to the temperature difference between the body and surroundings. Formula: dT/dt = -k(T - T₀), where k is cooling constant and T₀ is ambient temperature.

Thermal Properties of Matter for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Thermal Properties of Matter JEE notes, Formulas, PYQs — Thermal Properties of Matter - Complete JEE Physics Notes

Heat and Temperature

Understanding the distinction between heat and temperature is fundamental to thermodynamics and thermal physics. This forms the basis for all concepts in thermal properties of matter.

1.1 Heat vs Temperature

Heat (Q)

Definition: Energy in transit due to temperature difference
SI Unit: Joule (J)
CGS Unit: Calorie (cal), 1 cal = 4.186 J
Nature: Extensive property (depends on amount)
Direction: Always flows from hot to cold
Measurement: Calorimeter

Temperature (T)

Definition: Measure of average kinetic energy of molecules
SI Unit: Kelvin (K)
Other Units: °C, °F
Nature: Intensive property (independent of amount)
Significance: Determines direction of heat flow
Measurement: Thermometer

1.2 Temperature Scales & Conversions

Important Temperature Conversions

Celsius to Kelvin:

\[K = °C + 273.15\]

Celsius to Fahrenheit:

°F = \frac{9}{5}°C + 32

General Relation (JEE Important):

\frac{°C}{5} = \frac{°F - 32}{9} = \frac{K - 273}{5}

Scale	Lower Fixed Point	Upper Fixed Point	Number of Divisions
Celsius (°C)	0°C (Ice point)	100°C (Steam point)	100
Fahrenheit (°F)	32°F	212°F	180
Kelvin (K)	273.15 K	373.15 K	100

💡 JEE Quick Trick

For same numerical reading on Celsius and Fahrenheit:

\[°C = °F = -40\]

This is the only temperature where Celsius and Fahrenheit scales coincide!

📝 Solved Example 1

Question: At what temperature will the Fahrenheit scale reading be double the Celsius scale reading?

Solution:

Given: °F = 2°C

Using conversion formula:

°F = \frac{9}{5}°C + 32

Substitute °F = 2°C:

2°C = \frac{9}{5}°C + 32

2°C - \frac{9}{5}°C = 32

\frac{10°C - 9°C}{5} = 32

\frac{°C}{5} = 32

°C = 160°C\] \[°F = 320°F

⚠️ Common Mistakes

Never use °C in thermodynamic equations - always convert to Kelvin
Temperature difference ΔT is same in Celsius and Kelvin (ΔK = Δ°C)
Absolute zero (0 K) = -273.15°C (impossible to achieve)
Heat and temperature are NOT the same thing

Thermal Expansion

When a substance is heated, its molecules vibrate with greater amplitude, causing an increase in the average separation between molecules. This results in expansion - one of the most important concepts for JEE numerical problems.

2.1 Types of Thermal Expansion

Linear Expansion

For solids with one dominant dimension (rods, wires)

\Delta L = \alpha L_0 \Delta T

where α = coefficient of linear expansion

L = L_0(1 + \alpha \Delta T)

Area Expansion

For thin sheets, plates (2D objects)

\Delta A = \beta A_0 \Delta T

where β = 2α

A = A_0(1 + \beta \Delta T)

Volume Expansion

For 3D objects and liquids

\Delta V = \gamma V_0 \Delta T

where γ = 3α

V = V_0(1 + \gamma \Delta T)

Relation Between Coefficients (Most Important)

\beta = 2\alpha \quad \text{and} \quad \gamma = 3\alpha

Therefore: α : β : γ = 1 : 2 : 3

2.2 Important Values of α (Linear Expansion Coefficient)

Material	α (×10⁻⁶ /°C)	JEE Importance
Aluminum	23	High
Brass	19	Medium
Copper	17	High
Steel/Iron	12	High
Glass (Pyrex)	3.2	High
Invar (Fe-Ni alloy)	0.9	Very High

💡 JEE Special Cases

1. Expansion of a Cavity/Hole:

A hole in a metal plate expands as if it were made of the same material.

\Delta A_{\text{hole}} = \beta A_0 \Delta T

2. Thermal Stress:

When expansion is prevented, thermal stress develops:

\text{Thermal Stress} = Y \alpha \Delta T

where Y = Young's modulus

3. Bimetallic Strip:

Two metals with different α bonded together - bends towards metal with lower α when heated

📝 Solved Example 2

Question: A steel rod of length 1 m at 20°C is heated to 120°C. Find the increase in length. (α for steel = 12 × 10⁻⁶ /°C)

Solution:

Given:

L₀ = 1 m
T₁ = 20°C, T₂ = 120°C
ΔT = 120 - 20 = 100°C
α = 12 × 10⁻⁶ /°C

Using formula:

\Delta L = \alpha L_0 \Delta T

\Delta L = 12 \times 10^{-6} \times 1 \times 100

\Delta L = 12 \times 10^{-4} \text{ m}

\Delta L = 1.2 \text{ mm}

2.3 Anomalous Expansion of Water

❄️ Water's Unique Behavior (JEE Favorite)

Normal liquids: Volume decreases continuously as temperature decreases

Water: Volume decreases from 100°C to 4°C, then INCREASES from 4°C to 0°C

Key Points:

Maximum density of water occurs at 4°C
Water at 4°C sinks to bottom of lakes in winter
Ice floats because density of ice < density of water
This property helps aquatic life survive in frozen lakes

Calorimetry

Calorimetry is the science of measuring heat. The principle of calorimetry states that in an isolated system, heat lost by hot body = heat gained by cold body. This is the foundation for 30% of JEE thermal problems.

3.1 Fundamental Principle of Calorimetry

Heat Exchange Principle

\text{Heat Lost} = \text{Heat Gained}

Or equivalently:

\sum Q = 0

(Algebraic sum of all heat changes = 0)

3.2 Heat Transfer Formula

For Temperature Change (No Phase Change)

Q = mc\Delta T

Where:

Q = Heat transferred (J)
m = Mass (kg)
c = Specific heat capacity (J/kg·K)
ΔT = Temperature change (K or °C)

Sign Convention:

Q > 0: Heat absorbed (heating)
Q < 0: Heat released (cooling)
ΔT = T_final - T_initial

📝 Solved Example 3 (JEE Main Type)

Question: 100 g of water at 80°C is mixed with 200 g of water at 20°C in an insulated container. Find the final temperature. (Specific heat of water = 4200 J/kg·K)

Solution:

Given:

m₁ = 100 g = 0.1 kg (hot water)
T₁ = 80°C
m₂ = 200 g = 0.2 kg (cold water)
T₂ = 20°C
c = 4200 J/kg·K (same for both)

Let final temperature = T

Applying heat exchange principle:

\text{Heat lost by hot water} = \text{Heat gained by cold water}

\[m_1 c (T_1 - T) = m_2 c (T - T_2)\]

(c cancels out)

\[0.1(80 - T) = 0.2(T - 20)\]

\[8 - 0.1T = 0.2T - 4\]

\[12 = 0.3T\]

\[T = 40°C\]

💡 Method of Mixtures - Quick Formula

For mixing two quantities of same substance at different temperatures:

T_{\text{final}} = \frac{m_1T_1 + m_2T_2}{m_1 + m_2}

This is weighted average formula - saves 30 seconds in exam!

3.3 Water Equivalent

Definition & Formula

Water Equivalent (W): Mass of water that would absorb or release the same amount of heat as the body for the same temperature change.

W = \frac{mc}{c_{\text{water}}} = \frac{mc}{4200}

Where c = specific heat of the body

Heat Capacity:

H = mc = W \times c_{\text{water}}

Unit: J/K or J/°C

Specific Heat Capacity

Specific heat capacity is a characteristic property of every substance that determines how much heat is required to raise its temperature. Understanding this concept is crucial for solving 40% of calorimetry problems in JEE.

4.1 Definition & Formula

Specific Heat Capacity (c)

Heat required to raise the temperature of 1 kg of substance by 1 Kelvin.

c = \frac{Q}{m\Delta T}

SI Unit:

J/(kg·K) or J/(kg·°C)

CGS Unit:

cal/(g·°C)

4.2 Important Specific Heat Values (Must Remember)

Substance	c (J/kg·K)	c (cal/g·°C)	JEE Frequency
Water (Liquid)	4200	1.00	Highest
Ice	2100	0.50	High
Steam	2010	0.48	High
Aluminum	900	0.215	Medium
Iron/Steel	450	0.107	High
Copper	390	0.093	High
Lead	130	0.031	Low

💡 Why Water Has Highest Specific Heat?

Water has strong hydrogen bonding between molecules
Large amount of energy required to break these bonds
Application: Water is best coolant in engines and radiators
Climate effect: Coastal areas have moderate temperatures
Water at 4200 J/kg·K is reference standard (c_water = 1 cal/g·°C by definition)

4.3 Molar Specific Heat

Definition

Heat required to raise the temperature of 1 mole of substance by 1 Kelvin.

C = \frac{Q}{n\Delta T}

where n = number of moles

Relation with specific heat:

C = c \times M

where M = molar mass

SI Unit:

J/(mol·K)

CGS Unit:

cal/(mol·°C)

📝 Solved Example 4

Question: How much heat is required to raise the temperature of 2 kg of water from 20°C to 80°C? (c_water = 4200 J/kg·K)

Solution:

Given:

m = 2 kg
T₁ = 20°C, T₂ = 80°C
ΔT = 80 - 20 = 60°C = 60 K
c = 4200 J/kg·K

Using formula:

Q = mc\Delta T

Q = 2 \times 4200 \times 60

Q = 504000 \text{ J}

Q = 504 \text{ kJ} = 120 \text{ kcal}

Latent Heat & Change of State

When a substance changes its state (solid ↔ liquid ↔ gas), heat is absorbed or released at constant temperature. This hidden heat is called latent heat - one of the most important concepts for JEE Advanced problems.

5.1 Latent Heat - Definition & Types

Latent Heat of Fusion (L_f)

Heat required to convert 1 kg solid to liquid at melting point

\[Q = mL_f\]

For Water/Ice:

L_f = 334 kJ/kg

= 80 cal/g

Melting point = 0°C = 273 K

Latent Heat of Vaporization (L_v)

Heat required to convert 1 kg liquid to gas at boiling point

\[Q = mL_v\]

For Water/Steam:

L_v = 2260 kJ/kg

= 540 cal/g

Boiling point = 100°C = 373 K

⚠️ Key Points (JEE Important)

Temperature remains constant during phase change
L_v is always greater than L_f (for same substance)
For water: L_v/L_f ≈ 7 (approximately)
Steam at 100°C causes more severe burns than water at 100°C
Latent heat depends on pressure (increases with pressure for most substances)

5.2 Heating Curve for Water

Note: Horizontal portions represent phase changes at constant temperature

Total Heat Calculation (Ice → Steam)

To convert m kg of ice at -T₁°C to steam at T₂°C (above 100°C):

Step 1: Heat ice from -T₁ to 0°C

Q_1 = m \times c_{\text{ice}} \times T_1 = m \times 2100 \times T_1

Step 2: Melt ice at 0°C

Q_2 = m \times L_f = m \times 334000

Step 3: Heat water from 0 to 100°C

Q_3 = m \times c_{\text{water}} \times 100 = m \times 420000

Step 4: Boil water at 100°C

Q_4 = m \times L_v = m \times 2260000

Step 5: Heat steam from 100 to T₂°C

Q_5 = m \times c_{\text{steam}} \times (T_2-100) = m \times 2010 \times (T_2-100)

Total Heat:

Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5

📝 Solved Example 5 (JEE Advanced Type)

Question: How much heat is required to convert 10 g of ice at -10°C to steam at 100°C?
(c_ice = 2100 J/kg·K, c_water = 4200 J/kg·K, L_f = 334 kJ/kg, L_v = 2260 kJ/kg)

Solution:

Given: m = 10 g = 0.01 kg

Q₁ (Ice -10°C → 0°C):

Q_1 = 0.01 \times 2100 \times 10 = 210 \text{ J}

Q₂ (Melting at 0°C):

Q_2 = 0.01 \times 334000 = 3340 \text{ J}

Q₃ (Water 0°C → 100°C):

Q_3 = 0.01 \times 4200 \times 100 = 4200 \text{ J}

Q₄ (Boiling at 100°C):

Q_4 = 0.01 \times 2260000 = 22600 \text{ J}

Q_{\text{total}} = 210 + 3340 + 4200 + 22600 = 30350 \text{ J} = 30.35 \text{ kJ}

💡 JEE Shortcut - Rough Estimation

For converting ice at 0°C to steam at 100°C (1 kg):

Q_melting = 334 kJ (≈ 300 kJ)

Q_heating = 420 kJ (≈ 400 kJ)

Q_boiling = 2260 kJ (≈ 2300 kJ)

Total ≈ 3000 kJ (3 MJ)

Use this for quick MCQ elimination!

Heat Transfer - Conduction & Convection

Heat can be transferred through three modes: conduction, convection, and radiation. Understanding heat transfer is crucial as it appears in 25-30% of JEE thermal problems, often combined with other concepts.

6.1 Thermal Conduction

Fourier's Law of Heat Conduction

Rate of heat flow through a material is proportional to area and temperature gradient.

\frac{dQ}{dt} = kA\frac{(T_1 - T_2)}{L}

Where:

dQ/dt = Rate of heat flow (W)
k = Thermal conductivity (W/m·K)
A = Cross-sectional area (m²)
L = Length/thickness (m)
T₁ - T₂ = Temperature difference (K)

Alternative form:

\frac{dQ}{dt} = \frac{A(T_1-T_2)}{L/k} = \frac{T_1-T_2}{R}

where R = L/(kA) = Thermal resistance

6.2 Thermal Conductivity Values

Material	k (W/m·K)	Category	Application
Silver	420	Best Conductor	Electrical contacts
Copper	385	Good Conductor	Cooking vessels, wires
Aluminum	205	Good Conductor	Heat sinks
Steel	50	Moderate	Construction
Water	0.6	Poor Conductor	-
Wood	0.1-0.2	Insulator	Handles
Air	0.024	Best Insulator	Double walls

6.3 Compound Bar (Series & Parallel)

Series Combination

Same heat flow rate through both rods

\frac{dQ}{dt} = \frac{A(T_1-T_3)}{L_1/k_1 + L_2/k_2}

Equivalent thermal resistance:

R_{eq} = R_1 + R_2

Junction temperature T₂:

T_2 = \frac{k_1T_3L_2 + k_2T_1L_1}{k_1L_2 + k_2L_1}

Parallel Combination

Same temperature difference across both rods

\frac{dQ}{dt} = \frac{dQ_1}{dt} + \frac{dQ_2}{dt}

Equivalent thermal resistance:

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}

Or equivalently:

k_{eq}A_{eq} = k_1A_1 + k_2A_2

📝 Solved Example 6 (JEE Main Type)

Question: A copper rod of length 1 m and area 10⁻⁴ m² connects two reservoirs at 100°C and 0°C. Find the rate of heat flow. (k_copper = 385 W/m·K)

Solution:

Given:

L = 1 m
A = 10⁻⁴ m²
T₁ = 100°C, T₂ = 0°C
k = 385 W/m·K

Using Fourier's law:

\frac{dQ}{dt} = kA\frac{(T_1-T_2)}{L}

\frac{dQ}{dt} = 385 \times 10^{-4} \times \frac{100-0}{1}

\frac{dQ}{dt} = 385 \times 10^{-4} \times 100

\frac{dQ}{dt} = 3.85 \text{ W}

6.4 Convection

What is Convection?

Heat transfer through actual movement of heated material (fluids - liquids and gases).

Natural Convection:

Due to density differences
Sea breeze, land breeze
Hot air rises, cold sinks

Forced Convection:

Due to external force
Fans, pumps
More efficient

⚠️ JEE Key Points

Conduction: Dominant in solids (requires medium)
Convection: Only in fluids (requires medium + movement)
Good conductors: Metals (free electrons)
Poor conductors: Non-metals, liquids, gases
Best insulator: Vacuum (no conduction, no convection)

Thermal Radiation

Thermal radiation is electromagnetic wave emission by all bodies above absolute zero. Unlike conduction and convection, radiation does NOT require a medium - this is how Sun's heat reaches Earth through vacuum.

7.1 Stefan-Boltzmann Law

Power Radiated by Black Body

Total power radiated by a perfect black body is proportional to fourth power of absolute temperature.

P = \sigma A T^4

Where:

P = Power radiated (W)
σ = Stefan's constant
= 5.67 × 10⁻⁸ W/m²·K⁴
A = Surface area (m²)
T = Absolute temperature (K)

For Gray Body:

P = e\sigma A T^4

where e = emissivity (0 ≤ e ≤ 1)

e = 1 (perfect black body)
e = 0 (perfect reflector)

Net Rate of Heat Exchange

When a body at temperature T is in surroundings at temperature T₀:

\frac{dQ}{dt} = e\sigma A(T^4 - T_0^4)

This accounts for both emission and absorption

7.2 Wien's Displacement Law

Wavelength of Maximum Emission

As temperature increases, the wavelength at which maximum radiation occurs shifts to shorter wavelengths.

\lambda_m T = b

where b = Wien's constant = 2.9 × 10⁻³ m·K

Applications:

Sun (T ≈ 6000 K): λ_m ≈ 500 nm (visible, yellowish)
Human body (T ≈ 310 K): λ_m ≈ 9400 nm (infrared)
Hot metal (T ≈ 1000 K): λ_m ≈ 2900 nm (red glow)

7.3 Kirchhoff's Law of Radiation

Good Absorber = Good Emitter

At thermal equilibrium, the ratio of emissive power to absorptive power is same for all bodies and equals that of a black body.

\frac{E}{a} = \text{constant} = E_b

where E_b = emissive power of black body

Since a = e for a body:

\text{Absorptivity} = \text{Emissivity}

JEE Key Point: Black surface is both best absorber and best emitter!

📝 Solved Example 7

Question: A sphere of radius 10 cm and emissivity 0.8 is at 727°C. Find the power radiated. (σ = 5.67 × 10⁻⁸ W/m²·K⁴)

Solution:

Given:

r = 10 cm = 0.1 m
e = 0.8
T = 727°C = 1000 K
σ = 5.67 × 10⁻⁸ W/m²·K⁴

Step 1: Calculate surface area

A = 4\pi r^2 = 4 \times 3.14 \times (0.1)^2 = 0.1256 \text{ m}^2

Step 2: Apply Stefan's law

P = e\sigma A T^4

P = 0.8 \times 5.67 \times 10^{-8} \times 0.1256 \times (1000)^4

P = 0.8 \times 5.67 \times 0.1256 \times 10^4

P = 5700 \text{ W} = 5.7 \text{ kW}

💡 JEE Special Applications

1. Why we wear white in summer?

White has low absorptivity (reflects), keeps us cool

2. Why cooking utensils are black from bottom?

Black has high absorptivity, absorbs more heat

3. Thermos flask

Silvered walls (low e) minimize radiation loss, vacuum prevents conduction/convection

4. Solar cooker

Black interior (high absorption), glass cover (greenhouse effect)

Newton's Law of Cooling

Newton's law of cooling describes how objects cool down when placed in a cooler environment. This is one of the most important topics for JEE numerical problems, especially in JEE Advanced.

8.1 Newton's Law Statement

Law Statement

The rate of cooling of a body is proportional to the temperature difference between the body and its surroundings.

\frac{dT}{dt} = -k(T - T_0)

Where:

dT/dt = Rate of cooling
T = Temperature of body
T₀ = Ambient temperature
k = Cooling constant (depends on surface area, nature)

Conditions:

Valid when (T - T₀) is small
Usually ΔT < 30°C
Forced convection absent
Body temperature uniform

8.2 Temperature-Time Relation

Exponential Cooling Formula

Solving the differential equation:

T = T_0 + (T_i - T_0)e^{-kt}

where T_i = initial temperature at t = 0

Alternative useful forms:

T - T_0 = (T_i - T_0)e^{-kt}

\ln\left(\frac{T - T_0}{T_i - T_0}\right) = -kt

8.3 Average Rate of Cooling

For Small Temperature Difference

When temperature change is small, average rate ≈ instantaneous rate:

\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)

This is the most commonly used JEE formula!

📝 Solved Example 8 (Classic JEE Problem)

Question: A body cools from 80°C to 70°C in 5 minutes. If room temperature is 30°C, how much time will it take to cool from 70°C to 60°C?

Solution:

Given:

T₀ = 30°C (room temperature)
First cooling: 80°C → 70°C in t₁ = 5 min
Second cooling: 70°C → 60°C in t₂ = ?

For first cooling (80°C → 70°C):

\frac{80-70}{5} = k\left(\frac{80+70}{2} - 30\right)

\frac{10}{5} = k(75 - 30)

\[2 = 45k\]

k = \frac{2}{45} \text{ min}^{-1}

For second cooling (70°C → 60°C):

\frac{70-60}{t_2} = k\left(\frac{70+60}{2} - 30\right)

\frac{10}{t_2} = \frac{2}{45}(65 - 30)

\frac{10}{t_2} = \frac{2}{45} \times 35

t_2 = \frac{10 \times 45}{70} = \frac{450}{70}

t_2 = 6.43 \text{ minutes} \approx 6.4 \text{ min}

💡 Quick Tricks for Newton's Cooling

1. For equal temperature drops:

Time increases as average temperature decreases

\frac{t_2}{t_1} = \frac{T_{avg,1} - T_0}{T_{avg,2} - T_0}

2. Half-life cooling:

Time for (T - T₀) to become half:

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

3. Temperature graph:

T vs t is exponential decay, approaches T₀ asymptotically (never reaches exactly)

⚠️ Common Mistakes in JEE

Using Newton's law for large temperature differences (ΔT > 30°C)
Forgetting that k depends on surface area and nature of surface
Using wrong average temperature in rate formula
Not converting temperature to Kelvin when using Stefan's law alongside
Assuming body reaches ambient temperature in finite time (it's asymptotic!)

📝 Previous Year Questions Analysis (2015-2024)

JEE Main Pattern

✓ Calorimetry & Mixtures: 35%
✓ Thermal Expansion: 25%
✓ Heat Transfer (Conduction): 20%
✓ Newton's Law of Cooling: 15%
✓ Radiation (Stefan's law): 5%

JEE Advanced Pattern

✓ Complex calorimetry: 30%
✓ Compound bar problems: 25%
✓ Newton's cooling + radiation: 20%
✓ Thermal stress: 15%
✓ Mixed concepts: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 10-18 marks (3-5 questions)

Difficulty Level: Medium to Hard

Time Required: 15-20 hours study

Year-wise Trend Analysis

Year	JEE Main Focus	JEE Advanced Focus
2024	Calorimetry mixtures, Newton's cooling	Compound bar, Radiation + cooling
2023	Thermal expansion, Latent heat	Multi-step calorimetry, Thermal stress
2022	Heat conduction, Stefan's law	Complex phase change, Wien's law
2021	Mixing problems, Expansion	Series-parallel bars, Cooling

Download Complete PYQ PDF (2015-2024) with Solutions

🎯 Practice Problem Set

Level 1: Basic (JEE Main Foundation)

Convert 37°C to Fahrenheit and Kelvin scales.
A steel rod of length 2 m is heated from 20°C to 120°C. Find increase in length. (α = 12 × 10⁻⁶ /°C)
How much heat is required to raise temperature of 500 g of water from 25°C to 75°C? (c = 4200 J/kg·K)
Find final temperature when 100 g of water at 80°C is mixed with 200 g at 20°C.
Calculate heat required to convert 50 g of ice at 0°C to water at 0°C. (L_f = 334 kJ/kg)
A copper rod (k = 385 W/m·K) of length 50 cm and area 10⁻⁴ m² has ends at 100°C and 0°C. Find heat flow rate.
A black body at 727°C has surface area 100 cm². Find power radiated. (σ = 5.67 × 10⁻⁸ W/m²·K⁴)
Using Wien's law, find wavelength of maximum emission from Sun at 6000 K. (b = 2.9 × 10⁻³ m·K)

Level 2: Intermediate (JEE Main Standard)

100 g of ice at -10°C is mixed with 200 g of water at 30°C. Find final temperature and composition. (c_ice = 2100 J/kg·K, L_f = 334 kJ/kg)
A steel rod and copper rod of same length and area are joined in series. Ends are at 100°C and 0°C. Find junction temperature. (k_steel = 50, k_copper = 385 W/m·K)
A body cools from 100°C to 80°C in 5 min. Room temp is 20°C. Find time to cool from 80°C to 60°C.
A pendulum clock keeping correct time at 20°C is taken to 40°C. Find seconds it gains/loses per day. (α = 12 × 10⁻⁶ /°C)
Find water equivalent of 500 g copper calorimeter. (c_copper = 390 J/kg·K, c_water = 4200 J/kg·K)
Two rods of same material, length L and 2L, area A and 2A are joined in parallel. Find equivalent thermal resistance.
A sphere of radius R at temperature T radiates power P. If radius is doubled keeping T same, find new power.
How much heat is needed to convert 20 g of ice at -20°C to steam at 120°C? (All specific heats and latent heats given)

Level 3: Advanced (JEE Advanced/Olympiad)

Three rods of equal length and area are joined to form an equilateral triangle. Two junctions are at 100°C and 0°C. Find temperature of third junction and heat currents. (Thermal conductivities: k, 2k, 3k)
A body at temperature T₁ is placed in surroundings at T₀. It cools to temperature T₂ in time t. Derive expression for cooling constant k using Stefan's law and Newton's law.
A metallic sphere of radius R at temperature T is kept in surroundings at T₀. Find rate of change of temperature, given density ρ, specific heat c, emissivity e.
Ice at 0°C is added to m g of water at T°C. Just enough ice is added to cool water to 0°C. Find mass of ice melted. Show that it's independent of initial temperature T.
A rod of length L, area A, coefficient α is heated from T₁ to T₂ with ends fixed. Find thermal stress developed. (Young's modulus = Y)
Two plates separated by distance d have temperatures T₁ and T₂. A thin plate of conductivity k and thickness t is placed between them. Find its temperature assuming steady state.
A body loses half its temperature difference with surroundings in 10 minutes. How much time will it take to lose 87.5% of the difference? (Use Newton's cooling)
Derive the relation between coefficient of linear expansion of a solid and coefficient of cubical expansion of its liquid form at melting point, given latent heat and change in volume on melting.

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📑 Quick Navigation - Thermal Properties of Matter

Thermal Properties of Matter JEE Main & Advanced 2025-26

Heat and Temperature

1.1 Heat vs Temperature

Heat (Q)

Temperature (T)

1.2 Temperature Scales & Conversions

Important Temperature Conversions

💡 JEE Quick Trick

📝 Solved Example 1

⚠️ Common Mistakes

Thermal Expansion

2.1 Types of Thermal Expansion

Linear Expansion

Area Expansion

Volume Expansion

Relation Between Coefficients (Most Important)

2.2 Important Values of α (Linear Expansion Coefficient)

💡 JEE Special Cases

📝 Solved Example 2

2.3 Anomalous Expansion of Water

❄️ Water's Unique Behavior (JEE Favorite)

Calorimetry

3.1 Fundamental Principle of Calorimetry

Heat Exchange Principle

3.2 Heat Transfer Formula

For Temperature Change (No Phase Change)

📝 Solved Example 3 (JEE Main Type)

💡 Method of Mixtures - Quick Formula

3.3 Water Equivalent

Definition & Formula

Specific Heat Capacity

4.1 Definition & Formula

Specific Heat Capacity (c)

4.2 Important Specific Heat Values (Must Remember)

💡 Why Water Has Highest Specific Heat?

4.3 Molar Specific Heat

Definition

📝 Solved Example 4

Latent Heat & Change of State

5.1 Latent Heat - Definition & Types

Latent Heat of Fusion (L_f)

Latent Heat of Vaporization (L_v)

⚠️ Key Points (JEE Important)

5.2 Heating Curve for Water

Total Heat Calculation (Ice → Steam)

📝 Solved Example 5 (JEE Advanced Type)

💡 JEE Shortcut - Rough Estimation

Heat Transfer - Conduction & Convection

6.1 Thermal Conduction

Fourier's Law of Heat Conduction

6.2 Thermal Conductivity Values

6.3 Compound Bar (Series & Parallel)

Series Combination

Parallel Combination

📝 Solved Example 6 (JEE Main Type)

6.4 Convection

What is Convection?

⚠️ JEE Key Points

Thermal Radiation

7.1 Stefan-Boltzmann Law

Power Radiated by Black Body

Net Rate of Heat Exchange

7.2 Wien's Displacement Law

Wavelength of Maximum Emission

7.3 Kirchhoff's Law of Radiation

Good Absorber = Good Emitter

📝 Solved Example 7

💡 JEE Special Applications

Newton's Law of Cooling

8.1 Newton's Law Statement

Law Statement

8.2 Temperature-Time Relation

Exponential Cooling Formula

8.3 Average Rate of Cooling

For Small Temperature Difference

📝 Solved Example 8 (Classic JEE Problem)

💡 Quick Tricks for Newton's Cooling

⚠️ Common Mistakes in JEE

📝 Previous Year Questions Analysis (2015-2024)