4.1 Heat & Energy

Kinetic Energy of particles (from their movement)
+
Potential Energy of particles (from their position relative to each other)

Heating an object increases its internal energy. This can either:

1. Increase temperature → particles move faster (↑ kinetic energy)

2. Change state → particles break bonds (↑ potential energy)

During a change of state, temperature stays constant.

When ice melts at 0°C:

• Energy is being added
• But temperature stays at 0°C
• Energy is used to break bonds between particles
• This increases potential energy, not kinetic energy
• Once all bonds are broken, temperature rises again

$$\Delta E = m \times c \times \Delta\theta$$

Where:
• $\Delta E$ = energy transferred (J)
• $m$ = mass (kg)
• $c$ = specific heat capacity (J/kg°C)
• $\Delta\theta$ = change in temperature (°C)

• High SHC (like water): Takes a lot of energy to heat up, also takes a long time to cool down

• Low SHC (like metals): Heats up very quickly, also cools down quickly

This is why water is used in heating systems - it stores lots of energy.

How much energy is needed to heat 2 kg of water from 20°C to 100°C?
(Specific heat capacity of water = 4,200 J/kg°C)

Step 1: Find the temperature change
$\Delta\theta = 100 - 20 = 80°C$

Step 2: Write the formula
$\Delta E = m \times c \times \Delta\theta$

Step 3: Substitute values
$\Delta E = 2 \times 4200 \times 80$

Step 4: Calculate
$\Delta E = 672,000$ J = 672 kJ

A 0.5 kg aluminium block absorbs 9,000 J of energy. What is the temperature rise?
(Specific heat capacity of aluminium = 900 J/kg°C)

Step 1: Rearrange the formula
$\Delta\theta = \frac{\Delta E}{m \times c}$

Step 2: Substitute values
$\Delta\theta = \frac{9000}{0.5 \times 900}$

Step 3: Calculate
$\Delta\theta = \frac{9000}{450} = 20°C$

$$E = m \times L$$

Where:
• $E$ = energy transferred (J)
• $m$ = mass (kg)
• $L$ = specific latent heat (J/kg)

"Latent" means "hidden". The energy is hidden because:

• Temperature doesn't change during state change
• Energy goes into breaking/forming bonds
• A thermometer won't detect this energy transfer

Note: Vaporisation requires much more energy than fusion because gas particles must completely overcome all attractive forces.

How much energy is needed to melt 0.5 kg of ice at 0°C?
(Specific latent heat of fusion for water = 334,000 J/kg)

Step 1: Write the formula
$E = m \times L$

Step 2: Substitute values
$E = 0.5 \times 334,000$

Step 3: Calculate
$E = 167,000$ J = 167 kJ

How much energy is needed to turn 2 kg of water at 100°C into steam?
(Specific latent heat of vaporisation = 2,260,000 J/kg)

Step 1: Write the formula
$E = m \times L$

Step 2: Substitute values
$E = 2 \times 2,260,000$

Step 3: Calculate
$E = 4,520,000$ J = 4,520 kJ or 4.52 MJ

How much energy is needed to heat 0.2 kg of ice from -10°C to water at 50°C?
(c for ice = 2,100 J/kg°C, c for water = 4,200 J/kg°C, L fusion = 334,000 J/kg)

Step 1: Heat ice from -10°C to 0°C
$E_1 = 0.2 \times 2100 \times 10 = 4,200$ J

Step 2: Melt ice at 0°C
$E_2 = 0.2 \times 334,000 = 66,800$ J

Step 3: Heat water from 0°C to 50°C
$E_3 = 0.2 \times 4200 \times 50 = 42,000$ J

Total: $E = 4,200 + 66,800 + 42,000 = 113,000$ J = 113 kJ