Applied Physics - 2nd Semester Comprehensive Homework Guide
Unit 1: Thermodynamics Fundamentals
What is Thermodynamics? Thermodynamics is the study of heat, energy, and their effects on matter and radiation.
Key Concepts:
TEMPERATURE: Measure of thermal energy
- Celsius: Water freezes at 0°C, boils at 100°C
- Kelvin: Absolute scale, 0K = -273°C (absolute zero)
- Fahrenheit: Used in some countries
Conversion: K = °C + 273.15
HEAT: Transfer of thermal energy between objects
- Not same as temperature!
- Temperature = measure of molecular motion
- Heat = transfer of energy due to temperature difference
THERMAL ENERGY: Total kinetic energy of particles
- Depends on both temperature and amount of substanceLaw of Thermodynamics:
FIRST LAW (Energy Conservation):
Q = ΔU + W
Where:
Q = Heat added to system
ΔU = Change in internal energy
W = Work done by system
Meaning: Energy cannot be created or destroyed;
it changes form or transfers between objects
Example: When heating water:
- Heat energy (Q) makes water molecules move faster
- Some energy increases temperature (ΔU)
- Some energy does work expanding the liquid (W)SECOND LAW (Entropy):
In any spontaneous process:
Entropy (disorder) of isolated system increases
Meaning:
- Heat flows from hot object to cold object
- Never spontaneously flows the other way
- Efficiency of heat engines is never 100%
- Everything tends toward disorder
Example:
- Hot tea cools down (entropy increases)
- Cold tea never spontaneously heats (violates 2nd law)Heat Capacity & Specific Heat:
Heat Capacity (C):
Energy needed to raise temperature of substance by 1°C
C = Q / ΔT
Units: J/°C or cal/°C
Specific Heat (c):
Heat capacity per unit mass
c = Q / (m × ΔT)
Units: J/(kg·°C) or cal/(g·°C)
Water specific heat: 4186 J/(kg·°C)
- Highest of common materials
- Why water heats/cools slowly
- Why oceans regulate climate
Calculation Example:
How much heat to raise 2 kg water by 10°C?
Q = m × c × ΔT
= 2 kg × 4186 J/(kg·°C) × 10°C
= 83,720 JUnit 2: Heat Transfer Mechanisms
Three Ways Heat Transfers:
1. CONDUCTION
Through solid material (no particle movement)
Example: Metal spoon gets hot in hot tea
Rate of heat transfer:
Q = (k × A × ΔT × t) / d
Where:
k = Thermal conductivity
A = Cross-sectional area
ΔT = Temperature difference
t = Time
d = Thickness
Good conductors: Copper, Aluminum, Silver
Poor conductors: Wood, Foam, Rubber
2. CONVECTION
Through fluid movement (liquid or gas)
Example: Heat rising from radiator in room
Process:
- Hot fluid becomes less dense
- Rises up
- Cool fluid sinks down
- Creates circulation (convection current)
Used in: Ovens, cooling systems, weather patterns
3. RADIATION
Through electromagnetic waves (no medium needed)
Example: Sun heating Earth (through vacuum of space)
Stefan-Boltzmann Law:
P = σ × A × T⁴
Where:
P = Power radiated
σ = Stefan-Boltzmann constant
A = Surface area
T = Absolute temperature (Kelvin)
Key insight: Power depends on T⁴
- Small temperature increase = huge increase in radiation
- Why hot objects cool quickly at firstUnit 3: Thermodynamic Processes
Types of Processes:
ISOTHERMAL (Constant Temperature)
- ΔT = 0 (no temperature change)
- Example: Melting ice at 0°C
- For ideal gas: PV = constant
- First law: Q = W (all heat becomes work)
ADIABATIC (No Heat Transfer)
- Q = 0 (no heat exchange with surroundings)
- Example: Rapid compression of gas
- First law: ΔU = -W (temperature changes)
- Used in: Diesel engines, expanding clouds
ISOCHORIC (Constant Volume)
- V = constant (no volume change)
- W = 0 (no work done)
- First law: Q = ΔU (heat changes temperature)
- Example: Heating gas in rigid container
ISOBARIC (Constant Pressure)
- P = constant
- Example: Heating gas in piston (allows expansion)
- First law: Q = ΔU + W
- Most common in natureUnit 4 & 5: Kinetic Theory & Thermal Properties
Kinetic Theory of Gases:
Main Ideas:
1. Gas consists of particles in random motion
2. Collisions are perfectly elastic
3. Volume of particles << volume of gas
4. Temperature proportional to average kinetic energy
Ideal Gas Law:
PV = nRT
Where:
P = Pressure (Pa)
V = Volume (m³)
n = Number of moles
R = Gas constant = 8.314 J/(mol·K)
T = Absolute temperature (K)
Example:
A balloon at 1 atm, 20°C contains 22.4L air (1 mole)
Find if heated to 40°C at constant pressure:
V₁/T₁ = V₂/T₂ (Charles's Law)
V₂ = V₁ × (T₂/T₁)
= 22.4 × (313K / 293K)
= 23.9 L
The balloon expands by 1.5 L!Phase Changes:
SOLID ↔ LIQUID (Melting/Freezing)
- Latent heat of fusion: Energy to change state
- Ice at 0°C needs 334,000 J/kg to become water
- Temperature remains constant during phase change
LIQUID ↔ GAS (Vaporization/Condensation)
- Latent heat of vaporization: Much larger
- Water at 100°C needs 2,260,000 J/kg to become steam
- This is why steam causes severe burns
Temperature vs Heat Graph:During phase change, temperature constant but energy added!
### Problem-Solving Examples
**Example 1: Heat Transfer Problem**Problem: A copper rod (k=400 W/m·K) is 0.5m long and has area 0.01 m². One end is at 100°C, other at 0°C. Find heat transfer rate.
Solution: Q/t = (k × A × ΔT) / d = (400 × 0.01 × 100) / 0.5 = 400 / 0.5 = 800 W
Interpretation: 800 joules per second flow through the rod
**Example 2: Phase Change Problem**Problem: How much heat needed to convert 2 kg of ice at -10°C to steam at 110°C? (c_ice = 2100, c_water = 4186, c_steam = 1900 J/kg·K, L_f = 334,000, L_v = 2,260,000 J/kg)
Step 1: Heat ice from -10°C to 0°C Q₁ = m × c_ice × ΔT = 2 × 2100 × 10 = 42,000 J
Step 2: Melt ice at 0°C Q₂ = m × L_f = 2 × 334,000 = 668,000 J
Step 3: Heat water from 0°C to 100°C Q₃ = m × c_water × ΔT = 2 × 4186 × 100 = 837,200 J
Step 4: Vaporize water at 100°C Q₄ = m × L_v = 2 × 2,260,000 = 4,520,000 J
Step 5: Heat steam from 100°C to 110°C Q₅ = m × c_steam × ΔT = 2 × 1900 × 10 = 38,000 J
Total: Q = 42,000 + 668,000 + 837,200 + 4,520,000 + 38,000 = 6,105,200 J ≈ 6.1 MJ
Interpretation: Massive energy needed! Mostly for phase changes
### Real-World Applications
**Heat Engines (Convert Heat to Work):**EXAMPLE: Car Engine
- High temperature combustion (hot reservoir)
- Cylinder walls cooled by water (cold reservoir)
- Expanding gases push piston (work output)
- Efficiency: Typically 20-35% (losses as heat)
EXAMPLE: Power Plant
- Fuel burns, heats water
- Steam drives turbines
- Electricity generated
- Cooling tower rejects waste heat
- Efficiency: Typically 35-40%
**Heat Pumps (Move Heat Against Gradient):**EXAMPLE: Air Conditioning
- Moves heat from cool inside to hot outside
- Requires work input (electricity)
- COP (Coefficient of Performance) typically 2-4
- 1 kW electricity moves 2-4 kW heat
EXAMPLE: Heat Pump Heater
- Extracts heat from outside air/ground
- Moves to inside for heating
- More efficient than resistance heating
- Even works in cold weather
### Download Physics Notes by Units
**Applied Physics Homework & Notes:**
- [Download PDF of unit-1](https://drive.google.com/uc?export=download&id=1Q6Mafb-Q2KQq6x_Fd_PtO7Xk8OiGQTWg)
- [Download PDF of unit-2 and assignment-1](https://drive.google.com/uc?export=download&id=1WO9unQcZ3ZwWe3TjIpjn-gjUlDzKj_5u)
- [Download PDF of Unit-3 and assignment-2](https://drive.google.com/uc?export=download&id=12vKGUCECTBcrGln8Zh-LvU1xfCGtNygd)
- [Download PDF of Unit-4&5 and assignment-4&5](https://drive.google.com/uc?export=download&id=1C2M71mp-__Vcrsx5FVUs-_s9XJTPMTHh)
Each PDF contains:
- Complete unit theory
- Worked examples
- Practice problems
- Assignments with solutions
### Key Formulas Summary
THERMODYNAMICS:
- Temperature conversion: K = °C + 273.15
- Heat capacity: C = Q / ΔT
- Specific heat: Q = m × c × ΔT
- Heat conduction: Q = (k × A × ΔT × t) / d
- First law: Q = ΔU + W
- Ideal gas law: PV = nRT
- Stefan-Boltzmann: P = σ × A × T⁴
### Study Tips for Success
1. **Understand concepts first**: Don't just memorize formulas
2. **Work through examples**: See how concepts apply
3. **Practice problems**: Start easy, progress to hard
4. **Use diagrams**: Draw phase diagrams, pressure-temperature graphs
5. **Connect to real life**: Think about applications
6. **Group study**: Discuss concepts with classmates
7. **Ask questions**: Clarify doubts immediately
8. **Revise regularly**: Re-read notes weekly
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**Applied Physics is challenging but fascinating. Master thermodynamics and you'll understand everything from engines to weather to refrigerators!**