Lesson 14: Thermodynamics I

Week 14 — Heat, temperature, and phase changes

🎯 Learning Objectives

Understand temperature scales and thermal equilibrium
Calculate heat transfer via conduction, convection, and radiation
Apply specific heat and latent heat concepts

🎯 Intuition First: Why Does Metal Feel Colder Than Wood?

If you touch a metal spoon and a wooden spoon at the same room temperature, the metal feels much colder. Why? Because metal conducts heat away from your finger much faster. Temperature tells you how hot something is; thermal conductivity tells you how fast heat flows.

💡 Key Distinction:

Temperature — measures average kinetic energy of molecules (hot vs. cold)
Heat — energy in transit due to temperature difference (flows from hot to cold)
Thermal energy — total internal energy stored in a substance

📋 Temperature Scales

Scale	Freezing (Water)	Boiling (Water)	Conversion
°Celsius	0°C	100°C	C = K − 273.15
Kelvin	273.15 K	373.15 K	K = C + 273.15
°Fahrenheit	32°F	212°F	F = 9C/5 + 32

📋 Specific Heat

Q = mcΔT

Where c = specific heat capacity (J/kg·°C)

Specific heat tells you how much energy is needed to raise 1 kg by 1°C. Water has a very high specific heat (4,186 J/kg·°C), which is why it takes forever to boil and why oceans moderate coastal climates.

📋 Latent Heat (Phase Changes)

Q = mL

L_f (fusion) = 334,000 J/kg for water (melting/freezing)

L_v (vaporization) = 2,260,000 J/kg for water (boiling/condensing)

During a phase change, temperature stays constant — all the energy goes into breaking or forming molecular bonds.

📝 Worked Example: Heating Ice to Steam

How much energy to convert 0.5 kg of ice at −10°C to steam at 110°C?

Step 1 — Heat ice from −10°C to 0°C:

Q₁ = mcΔT = 0.5 × 2100 × 10 = 10,500 J

Step 2 — Melt ice at 0°C:

Q₂ = mL_f = 0.5 × 334,000 = 167,000 J

Step 3 — Heat water from 0°C to 100°C:

Q₃ = mcΔT = 0.5 × 4186 × 100 = 209,300 J

Step 4 — Boil water at 100°C:

Q₄ = mL_v = 0.5 × 2,260,000 = 1,130,000 J

Step 5 — Heat steam from 100°C to 110°C:

Q₅ = mcΔT = 0.5 × 2010 × 10 = 10,050 J

Total:

Q_total = 10,500 + 167,000 + 209,300 + 1,130,000 + 10,050 = 1,526,850 J ≈ 1.53 MJ

📋 Heat Transfer Methods

Method	Mechanism	Formula
Conduction	Direct contact, molecular collisions	P = kAΔT/L
Convection	Moving fluid carries heat	Empirical (depends on flow)
Radiation	Electromagnetic waves	P = σAεT⁴

✅ Check Your Understanding

Q1: Why does sweating cool you down?

Click to reveal answer

When water evaporates from your skin, it requires energy (latent heat of vaporization = 2,260 kJ/kg). That energy comes from your body, so you lose thermal energy. This is why sweating is very effective — each gram of sweat removes 2,260 J of heat!

Q2 (mini-problem): A 3-kg aluminum block (c = 900 J/kg·°C) at 100°C is dropped into 2 kg of water at 20°C. What is the final temperature? (Assume no heat loss.)

Click to reveal answer

Heat lost by aluminum = Heat gained by water

3(900)(100 − T_f) = 2(4186)(T_f − 20)

270,000 − 2700T_f = 8372T_f − 167,440

437,440 = 11,072T_f

T_f = 39.5°C

Answer: Final temperature ≈ 39.5°C