10.6 – Latent Heat & Specific Latent Heat

Thermal Energy and Change in State

  • The transfer of thermal energy into (or out of) a body can cause a change in the physical state of the body.

Latent Heat

  • Melting / solidification and boiling / condensation are processes involving a change in state without a change in temperature.
  • During melting and boiling, energy absorbed
    • is used to overcome intermolecular forces
    • increases the internal energy of the material
  • During boiling, extra work is done against atmospheric pressure. Hence, energy involved in boiling/ condensation is greater than energy involved in melting/ solidification.
  • During boiling, melting and freezing, a substance undergoes a change in state without a change in temperature.
Definition: Latent Heat
Latent heat is the amount of thermal energy needed to change a substance from solid to liquid or from liquid to gas, without any change in temperature.

latent heat = Q

Units of latent heat are the same as that of heat i.e. joules (J).

We normally consider the latent heat per unit mass (kg) of matter. This is called Specific Latent Heat.

When referring to the solid ⇔ liquid phase change we use the term Specific Latent Heat of Fusion.

Definition: Specific Latent Heat of Fusion
Specific Latent Heat of Fusion is the amount of thermal energy needed to change unit mass of a substance from solid to liquid, without any change in temperature

If = Q / m

OR

Q = lf × m

Units: J kg⁻¹

 

When referring to the liquid ⇔ gas phase change we use the term Specific Latent Heat of Vaporisation.

Definition: Specific Latent Heat of Vaporisation
Specific Latent Heat of Vaporisation is the amount of thermal energy needed to change unit mass of a substance from liquid to gas, without any change in temperature

lv = Q / m

OR

Q = lv × m

Units: J kg⁻¹

 

  • Specific latent heat is more commonly used because it is the fundamental property of a substance or material.
  • Thermal energy (Q) gained/lost in the change of state is related to the mass of the substance (m)according to the equation: Q = m l
  • Each substance would have a specific latent heat of fusion and a specific latent heat of vaporisation.
  • A joulemeter is a device to measure the energy used in joules (J) in an electrical circuit or appliance.
Example 9
Water is vaporised from a beaker using a heater attached to a joulemeter (a device to measure the energy used in joules (J) in an electrical circuit or appliance).

mass of water vaporised = 20 g
joulemeter reading = 45 000 J

Calculate the specific latent heat of vaporisation of water.

20 g = 0.020 kg

Q = lv × m

45000 = lv × 0.020

lv = 2.25 × 106 J kg-1

Problem solving approach

 

Example 10
A 2.0 kW kettle containing boiling water is placed on a balance. It is left there and continues to boil for 5.0 minutes. The balance reading changes by 0.20 kg.

Determine the specific latent heat of vaporisation of water.

Energy supplied by kettle = Energy gained by water
(to become a gas) 

P t = m lv 

(2000 W)(5.0 x 60 s) = (0.20 kg) lv

lv = 3.0×106 J kg-1 
Example 12
A glass of lemonade with a mass of 0.20 kg is initially at 30 °C.
0.050 kg of ice at 0 °C is added to the lemonade and all of it melts.

  • specific heat capacity of water = 4200 J (kg °C)–1
  • specific latent heat of fusion of water = 3.34 × 105 J kg–1

Determine the final temperature of the lemonade.

Let θ be the final temperature (when thermal equilibrium is reached).

Assume that lemonade has the same specific heat capacity as water.

Word equation:
energy lost        =  energy gained by ice to melt + energy gained by water
by lemonade

Q = m1 c Δθ = m2 lf + m2 c Δθ

(0.20)(4200)(30 – θ) = (0.050)( 3.34 × 105) + (0.050)(4200)(θ – 0) 

(840)(30 – θ) = 16700 + 210 θ 

25200 – 840 θ = 16700 + 210 θ 

θ = (25200 – 16700) / (210 + 840)

θ = 8.095 

θ ≈ 8.1 °C (2 sig. fig.)
Example 13
10 g of ice are added to 200 g of water are mixed together in an insulated chamber.

The ice is at –20.0 °C while the water is at 30.0 °C.

Given:

specific heat capacity of water = 4200 J (kg °C)-1
specific heat capacity of ice = 2050 J (kg °C)-1
specific latent heat of fusion of water = 3.34 × 105 J kg-1

Find the final temperature of the mixture.

Let  θ °C be the final temperature (assume it is above 0 °C)

Where:

Q1 is the energy needed to warm the ice from –20 °C to 0 °C.

Q2 is the energy needed to melt the ice.

Q3 is the energy needed to warm the melted water from 0 °C to θ °C.

Q4 is the energy lost by 200 g of water as it cools from 30 °C to θ °C.

From conservation of energy:

Q4 = Q1 + Q2 + Q3

m2 c2 (30 – θ) = m1 c (0 – (–20)) + m1 If + m1 c2 (θ – 0)

Convert all masses to “kg”.

(0.20 kg)(4200 J (kg °C)-1) (30 °C – θ)     = (0.01 kg)(2050 J (kg °C)-1 )(20°C) + (0.01 kg)(3.34 × 105 J kg-1) + (0.01 kg)( 4200 J (kg °C)-1) θ

840(30 – θ)   =   410 + 3340 + 42 θ

θ (42 + 840)  =   25200 – 410 – 3340

θ = 24.32

θ ≈ 24 °C (2 sig. fig.)

 

Experimental Determination of Specific Latent Heats (ENRICHMENT)
Experimental Determination of Specific Latent Heat of Fusion of Water

Procedure

  1. Set up the apparatus as in Set-up A
  2. Measure initial mass of water
  3. Switch on the heater
  4. After a certain time interval, switch off the heater and measure final mass of water.
  5. Record reading on joulemeter attached to heat supply

Measurements & Results

  • Mass m of substance melted (difference in masses of water)
  • Total thermal energy supplied to the substance

Q = P t = m Lf

Sources of Error

  • There is gain of thermal energy from the surroundings.

Possible Improvements to the Set-up

Set-up B can be conducted at the same time as running experiment  Set-up A. This will show how much ice melt is due to gaining heat from the surroundings (rather than melted as a direct result of the heat energy supplied by the heater).
Experimental Determination of Specific Latent Heat of Vaporisation of Water

Procedure

  1. The water is set boiling with a heater.
  2. The joulemeter reading and mass of water is recorded after some bubbles are formed
  3. After a certain time interval, switch off the heater and measure the final mass of water and final joulemeter reading

Measurements & Results

  • Mass m of substance vaporised (difference in masses of water)
  • Total thermal energy supplied to the substance

Q = P t = m Lv

Sources of Error

  • There is gain of thermal energy from the surroundings.

Possible Improvements to the Set-up

  • Use a control to estimate the rate of lass of water mass due to evaporation.
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