### Specific Latent Heat

- The specific latent heat of a substance is the amount of heat requires to change the phase of 1 kg of substance at a constant temperature.
- Specific latent heat is measured in J/kg, if energy is measured in J and mass in kg.For example, specific latent heat of ice is 334000J/kg means 334000 J of energy is needed to convert 1kg of water into ice or vice versa.

Formula: - The specific latent heat of vaporization is the heat needed to change 1 kg of a liquid at its boiling point into vapour without a change in temperature.
- The specific latent heat of fusion is the heat needed to change 1 kg of a solid at its melting point into a liquid, without a change in temperature.
- If any solid is to become a liquid, it must gain the necessary latent heat. Equally, if a liquid is to change back into a solid, it must lose this latent heat.

**Example 1**:

How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K).]

**Answer**:

m = 2kg

Specific latent heat of fusion of water, L = 334 000 J/kg

specific heat capacity of water = 4200 J/(kg K)

Energy needed to melt 2kg of ice,

Q_{1} = mL = (2)(334000) = 668000J

Energy needed to change the temperature from 0°C to °C.

Q_{2} = mcθ = (2)(4200)(20 – 0) = 168000J

Total energy needed = Q_{1} + Q_{2} = 668000 + 168000 = 836000J

**Example 2**:

Starting at 20°C, how much heat is required to heat 0.3 kg of aluminum to its melting point and then to convert it all to liquid? [Specific heat capacity of aluminium = 900J kg^{-1} °C^{-1}; Specific latent heat of aluminium = 321,000 Jkg^{-1}, Melting point of aluminium = 660°C]

**Answer**:

m = 0.3kg

Specific latent heat of fusion of aluminium, L = 321 000 J/kg

specific heat capacity of aluminium = 900 J/(kg K)

Energy needed to increase the temperature from 20°C to 660°C

Q_{1} = mcθ = (0.3)(900)(660 – 20) = 172,800J

Energy needed to melt 0.3kg of aluminium,

Q_{2} = mL = (0.3)(321000) = 96,300J

Total energy needed = Q_{1} + Q_{2} = 172,800 + 96,300 = 269,100J

**Example 3**:

How much heat must be removed by a refrigerator from 2 kg of water at 70 °C to convert it to ice cubes at -11°C? [Specific heat capacity of water = 4200J kg^{-1} °C^{-1}; Specific latent heat of fusion of ice = 334,000 Jkg^{-1}, specific heat capacity of ice = 2100 J/(kg K)]

**Answer**:

m = 2kg

Specific latent heat of fusion of water, L = 334,000 J/kg

Specific heat capacity of water, c_{w} = 4,200 J/(kg K)

Specific heat capacity of ice, c_{i} = 2,100 J/(kg K)

Energy to be removed to reduce the temperature from 70°C to 0°C (Freezing point of water)

Q_{1} = mcθ = (2)(4200)(70 – 0) = 588,000J

Energy needed to freeze 2kg of water,

Q_{2} = mL = (2)(334,000) = 668,000J

Energy to be removed to reduce the temperature from 0°C to -11°C

Q_{3} = mcθ = (2)(2100)(0 – (-11)) = 46,200J

Total energy needed = Q_{1} + Q_{2} + Q_{3 }= 588,000 + 668,000 = 46,200J = 1,302,200J