**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