4.3.4 Specific Latent Heat

  1. 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.
  2. 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:
  3. 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.
  4. 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.
  5. 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.

Measuring the Specific Latent Heat of Fusing of Ice

  1. Figure above shows the apparatus setup to determine the specific latent heat of fusion of ice. Some ice at 0 °C is heated by a small electric heater which is left switched on for several minutes. 
  2. Some of the ice melts to form water which runs down through the funnel and is collected in the beaker. 
  3. The mass of ice (m) melted is found by measuring the mass of water collected.
  4. If the power of the heater is P and the time taken to heat the ice = t, then the thermal energy supplied by the heater = thermal energy used to melt ice = Pt.
    Therefore, the specific latent heat of fusion of ice

Precaution Steps:

  1. The heating element of the heater must fully immerse in ice so that all the heat generated is absorbed by the ice.
  2. A control set is needed to estimate the amount of mass of ice melted by the heat from the surrounding.

Note:

  1. The heat received by ice is less than the calculated value Pt as some heat is lost to the surrounding. This will result in the value of l obtained from the calculation to be slightly higher than the standard value.
  2. If impurity is present in water, the melting point of the water will be lower than normal.

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,
Q1 = mL = (2)(334000) = 668000J

Energy needed to change the temperature from 0°C to °C.
Q2 = mcθ  = (2)(4200)(20 - 0) = 168000J

Total energy needed = Q1 + Q2 = 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
Q1 = mcθ  = (0.3)(900)(660 - 20) = 172,800J

Energy needed to melt 0.3kg of aluminium,
Q2 = mL = (0.3)(321000) = 96,300J

Total energy needed = Q1 + Q2 = 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, cw = 4,200 J/(kg K)
Specific heat capacity of ice, ci = 2,100 J/(kg K)

Energy to be removed to reduce the temperature from 70°C to 0°C (Freezing point of water)
Q1 = mcθ  = (2)(4200)(70 - 0) = 588,000J

Energy needed to freeze 2kg of water,
Q2 = mL = (2)(334,000) = 668,000J

Energy to be removed to reduce the temperature from 0°C to -11°C
Q3 = mcθ  = (2)(2100)(0 - (-11)) = 46,200J

Total energy needed = Q1 + Q2  + Q= 588,000 + 668,000 = 46,200J = 1,302,200J