Heat Capacity and Latent Heat

Overview

This page focuses on how thermal energy changes the temperature or state of a substance.

Two major possibilities when energy is supplied:

1. Temperature changes
   Use:

   $$Q=C\Delta T$$

   or

   $$Q=mc\Delta T$$

2. State changes at constant temperature
   Use:

   $$Q=ml$$

This page supports:

• Thermal Physics A

Definition

Heat capacity describes energy needed to change temperature. Latent heat describes energy needed to change state without temperature change.

Why It Matters

This topic explains why water heats slowly, why metals heat quickly, why melting and boiling occur at constant temperature, and how calorimetry questions are really conservation-of-energy problems.

Key Representations

Heat, Temperature and Internal Energy

Heat

Heat is energy transferred due to a temperature difference.

Unit:

$$\mathrm{J}$$

Temperature

Temperature measures the degree of hotness and determines direction of heat flow.

Internal Energy

Internal energy is the total microscopic energy of particles:

• random kinetic energy
• intermolecular potential energy

Supplying heat usually increases internal energy.

Heat Capacity

Definition

Heat capacity (C) of an object is the thermal energy required to raise its temperature by (1\ \mathrm{K}) (or (1^\circ\mathrm{C})).

Formula

$$Q=C\Delta T$$

Where:

• (Q) = thermal energy supplied
• (C) = heat capacity
• (\Delta T) = temperature rise

Unit

$$\mathrm{J}{\mathrm{K}}^{-1}$$

Notes

Heat capacity depends on:

• mass of object
• material of object

A larger object usually has a larger heat capacity.

Specific Heat Capacity

Definition

Specific heat capacity (c) is the thermal energy required to raise the temperature of 1 kg of a substance by (1\ \mathrm{K}).

Formula

$$Q=mc\Delta T$$

Where:

• (m) = mass
• (c) = specific heat capacity

Unit

$$\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$$

Meaning of Large or Small Specific Heat Capacity

Large (c)

Needs more energy for same temperature rise.

Examples:

• water

Implications:

• warms slowly
• cools slowly
• useful in engine cooling and climate moderation

Small (c)

Needs less energy for same temperature rise.

Examples:

• many metals

Implications:

• heats quickly
• cools quickly

Comparing Heat Capacity and Specific Heat Capacity

Heat Capacity (C)

Applies to entire object.

Specific Heat Capacity (c)

Property of material.

Relation:

$$C=mc$$

Worked Example 1

A block has heat capacity:

$$C=800\mathrm{J}{\mathrm{K}}^{-1}$$

Find heat needed to raise temperature by (15\ \mathrm{K}).

$$Q=C\Delta T=800(15)=1.20\times 10^4\mathrm{J}$$

Worked Example 2

A (2.0\ \mathrm{kg}) mass of water is heated from (20^\circ\mathrm{C}) to (60^\circ\mathrm{C}).

Take:

$$c=4200\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$$ $$Q=mc\Delta T$$ $$Q=2.0(4200)(40)=3.36\times 10^5\mathrm{J}$$

Calorimetry and Mixing Problems

Core Principle

For an insulated system:

$$\text{Heat lost}=\text{Heat gained}$$

This follows conservation of energy.

Typical Mixing Setup

Hot object placed in cooler water.

Final temperature becomes common equilibrium temperature.

Use:

$$m_1{c}_1(T_1-T_f)=m_2{c}_2(T_f-T_2)$$

where:

• hot object cools
• cold object warms

Worked Example 3

A (0.50\ \mathrm{kg}) copper block at (100^\circ\mathrm{C}) is placed in (1.0\ \mathrm{kg}) water at (20^\circ\mathrm{C}).

Take:

• (c_{\text{Cu}}=390)
• (c_w=4200)

Find final temperature (T_f).

$$0.50(390)(100-T_f)=1.0(4200)(T_f-20)$$ $$195(100-T_f)=4200(T_f-20)$$ $$19500-195T_f=4200T_f-84000$$ $$103500=4395T_f$$ $$T_f\approx 23.5^{\circ }\mathrm{C}$$

Water temperature changes much less because water has large thermal capacity.

Important Assumptions in Calorimetry

Usually assume:

• no heat loss to surroundings
• no evaporation
• container heat capacity negligible (unless given)
• final equilibrium reached

If calorimeter has heat capacity (C), include:

$$Q=C\Delta T$$

Latent Heat

Meaning

During some heating processes, temperature remains constant while energy is still absorbed.

This energy changes molecular arrangement instead of increasing average kinetic energy.

Specific Latent Heat

Formula

$$Q=ml$$

Where:

• (m) = mass
• (l) = specific latent heat

Unit

$$\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$$

Types of Latent Heat

Specific Latent Heat of Fusion (l_f)

Energy required to change:

• solid (\rightarrow) liquid

at constant temperature.

Specific Latent Heat of Vaporisation (l_v)

Energy required to change:

• liquid (\rightarrow) gas

at constant temperature.

Usually:

$$l_v>l_f$$

because particles separate much more fully.

Worked Example 4

Energy to melt (0.30\ \mathrm{kg}) ice at (0^\circ\mathrm{C}):

Take:

$$l_f=3.34\times 10^5$$ $$Q=ml=0.30(3.34\times 10^5)$$ $$Q=1.00\times 10^5\mathrm{J}$$

Worked Example 5

Energy to boil away (0.50\ \mathrm{kg}) water at (100^\circ\mathrm{C}):

Take:

$$l_v=2.26\times 10^6$$ $$Q=0.50(2.26\times 10^6)$$ $$Q=1.13\times 10^6\mathrm{J}$$

Heating Curves

A heating curve shows temperature against time or energy supplied.

Typical stages:

1. solid warms
2. melting plateau
3. liquid warms
4. boiling plateau
5. gas warms

Figure: A heating curve for a pure substance, showing sloping regions for temperature rise and flat regions for phase change.

Read the graph section by section: sloping parts use (Q=mc\Delta T), while flat parts use (Q=ml).

Interpretation of Sloping Sections

Temperature rises:

Use:

$$Q=mc\Delta T$$

Energy increases average kinetic energy.

Interpretation of Flat Sections

Temperature constant:

Use:

$$Q=ml$$

Energy increases separation of particles or intermolecular potential energy.

Multi-Step Energy Problems

Sometimes combine formulas.

Example: Ice at (-10^\circ\mathrm{C}) to steam at (100^\circ\mathrm{C}):

1. Warm ice
2. Melt ice
3. Warm water
4. Boil water

Use the correct equation for each stage.

Worked Example 6

Find energy to convert (0.20\ \mathrm{kg}) ice at (0^\circ\mathrm{C}) to water at (25^\circ\mathrm{C}).

Take:

• (l_f = 3.34\times10^5)
• (c_w=4200)

Step 1: Melt ice

$$Q_1=ml=0.20(3.34\times 10^5)=66800$$

Step 2: Heat water

$$Q_2=mc\Delta T=0.20(4200)(25)=21000$$

Total

$$Q=87800\mathrm{J}$$

Links

• Thermal Physics A
• Thermal Measurement and Scales
• Thermal Practicals
• Thermal Physics A Common Exam Traps
• Work, Energy and Power

Summary

The key skill is deciding whether the energy supplied is:

• raising temperature
• changing state
• or doing both in stages

Once that is clear, the formula choice usually becomes straightforward.