Specific Heat abd Latent Heat

1. Thermal Equilibrium

Definition

Thermal equilibrium occurs when:

  • Two objects reach the same temperature.

  • Heat transfer stops.

Heat naturally flows:

  • From warmer object → cooler object

When temperatures become equal:

  • No net heat transfer occurs.

Example

Ice placed in soda:

  • Soda loses heat.

  • Ice gains heat.

  • Eventually both reach the same temperature.

This is:

Thermal Equilibrium

2. Zeroth Law of Thermodynamics

Definition

If:

  • Object A is in thermal equilibrium with B

  • Object C is in thermal equilibrium with B

Then:

  • A and C are also in thermal equilibrium.

This means:

  • They all have the same temperature.

Importance

The zeroth law:

  • Defines temperature

  • Makes thermometers possible

3. Thermometers

How They Work

A thermometer:

  • Reaches thermal equilibrium with an object.

  • Measures temperature from this equilibrium.

Example

Thermometer under tongue:

  • Heat transfers until temperatures equalize.

  • Thermometer reading becomes body temperature.

4. Energy Units

Joule (J)

SI unit for energy.

1 joule:

  • 1 Newton × meter

5. Specific Heat Capacity

Definition

Specific heat is:

  • The amount of heat needed to raise 1 gram of a substance by 1°C.

Units:

  • J/g°C

Formula

C=QmΔTC=\frac{Q}{m\Delta T}C=mΔTQ​

Where:

  • CCC = specific heat

  • QQQ = heat energy

  • mmm = mass

  • ΔT=Tf−Ti\Delta T = T_f - T_iΔT=Tf​−Ti​

Rearranged Formula

Q=mC(Tf−Ti)Q=mC(T_f-T_i)Q=mC(Tf​−Ti​)

Used to calculate heat gained or lost.

6. Specific Heat of Common Substances

Substance

Specific Heat (J/g°C)

Gold

0.13

Copper

0.39

Ice

2.05

Steam

2.08

Liquid Water

4.18

7. Low vs High Specific Heat

Low Specific Heat

  • Heats quickly

  • Cools quickly

Examples:

  • Gold

  • Copper

Applications:

  • Jewelry

  • Metal wires

High Specific Heat

  • Requires more energy to heat up

  • Changes temperature slowly

Example:

  • Water

8. Why Water Is Important

Water has:

  • Very high specific heat

Effects:

  • Lakes change temperature slowly

  • Earth’s climate stays more stable

  • Animals survive winter in ponds

  • Water cools us in summer

9. Specific Heat Example

Problem

Heat required to raise:

  • 120 g water

  • From 52°C to 85°C

Given:

  • C=4.18 J/g°CC = 4.18 \, J/g°CC=4.18J/g°C

Step 1: Temperature Change

ΔT=85−52=33∘C\Delta T = 85-52 = 33^\circ CΔT=85−52=33∘C

Step 2: Use Formula

Q=(120)(4.18)(33)Q=(120)(4.18)(33)Q=(120)(4.18)(33)

Answer

Q≈17000 JQ\approx17000\ JQ≈17000 J

10. Positive and Negative Q

Sign

Meaning

+Q

Heat absorbed

−Q

Heat released

11. Latent Heat

Definition

Latent heat:

  • Heat absorbed or released during a phase change.

  • Temperature does NOT change.

Energy changes:

  • Potential energy changes

  • Kinetic energy stays constant

12. Phase Changes

Change

Process

Solid → Liquid

Melting

Liquid → Gas

Boiling

Gas → Liquid

Condensation

Liquid → Solid

Freezing

13. Heating Curve

A heating curve shows:

  • Temperature changes during heating.

Phase A

  • Solid warms up.

  • Temperature rises.

Phase B

  • Melting occurs.

  • Temperature stays constant.

  • Heat breaks bonds.

Phase C

  • Liquid warms up.

  • Temperature rises.

Phase D

  • Boiling occurs.

  • Temperature constant again.

Phase E

  • Gas warms up.

  • Temperature rises.

14. Cooling Curve

Cooling curve is the reverse process.

Phase F

  • Gas cools.

Phase G

  • Condensation.

Phase H

  • Liquid cools.

Phase I

  • Freezing.

Phase J

  • Solid cools.

15. Important Facts About Curves

Sloped Lines

  • Temperature changes.

Flat Plateaus

  • Phase changes occur.

  • Temperature stays constant.

16. Heat of Fusion

Definition

Heat required to melt a solid.

For water:

  • Hf=334 J/gH_f = 334\ J/gHf​=334 J/g

17. Multi-Step Heating Problem

Problem

Convert:

  • 50 g ice at −5°C
    → liquid water at 65°C

Step 1: Heat Ice to 0°C

Q=(50)(2.08)(0−(−5))Q=(50)(2.08)(0-(-5))Q=(50)(2.08)(0−(−5))

Answer:

Q=520 JQ=520\ JQ=520 J

Step 2: Melt Ice

Q=Hfm=(334)(50)Q=H_fm=(334)(50)Q=Hf​m=(334)(50)

Answer:

Q=16700 JQ=16700\ JQ=16700 J

Step 3: Heat Water to 65°C

Q=(50)(4.18)(65−0)Q=(50)(4.18)(65-0)Q=(50)(4.18)(65−0)

Answer:

Q=13585 JQ=13585\ JQ=13585 J

Step 4: Total Heat

Qtotal=520+16700+13585Q_{total}=520+16700+13585Qtotal​=520+16700+13585

Answer:

Qtotal=30805 JQ_{total}=30805\ JQtotal​=30805 J

18. Core Concepts Summary

  • Thermal equilibrium = same temperature.

  • Zeroth law explains temperature comparison.

  • Specific heat measures resistance to temperature change.

  • Water has high specific heat.

  • Latent heat occurs during phase changes.

  • Flat parts of heating curves = phase changes.

  • Sloped parts = temperature changes.

  • Melting/freezing happen at same temperature.

  • Boiling/condensation happen at same temperature.