Back to: PHYSICS SS2

**Welcome to class! **

In today’s class, we will be talking about latent heat. Enjoy the class!

**Latent Heat**

**Definition of Latent Heat**

Latent heat is the quantity of heat energy that is required to change a substance from one state to another at constant temperature and pressure. The unit of latent heat is Joule (J), and the symbol used for latent heat is Q.

**The formula of Latent Heat Capacity**

The quantity of heat that is required to change a substance from one state to another is directly proportional to the mass of the substance.

Mathematically:

Quantity of heat energy mass of the substance

Introducing the constant of proportionality: Q = L x m

L is the constant of proportionality. It is called the specific latent heat of the substance.

M is the mass of the substance measured in a kilogram.

Q is the quantity of heat measured in Joule.

**Factors that determine the quantity of heat required to change the state of a substance**

- The nature of the substance.
- The mass of the substance.
- The quantity of energy supplied.
- The volume of the substance
- The area of the substance

**Specific Latent Heat of a Substance**

Specific latent heat of a substance is the quantity of heat energy that is required to change 1kg mass of the substance from one state to another at a constant temperature.

**Specific Latent Heat of Fusion of Ice L _{ice}:**

Specific latent heat of fusion of ice is the quantity of heat energy that is required to change 1kg mass of ice block into a liquid at constant temperature Or the melting point of the ice block. It is a scalar quantity.

Specific Latent Heat of Fusion of Ice L ice = 336 J/g

Formula Of Specific Latent Heat Of Fusion Of Ice:

Quantity of Heat energy Q = mass of ice block x Specific latent heat of ice block

Q = m x L_{ice}

The specific latent beat of ice L_{ice} = Q ÷ m

**WORKED EXAMPLES**

(1) Calculate the quantity of heat energy that is required to change 25.6 kg mass of ice to liquid at a constant temperature.

SOLUTION

Data given in the question:

Mass of ice block m = 25.6 kg = 25.6 x 1000 = 25600g,

L_{ice} = 336J/g,

Q = Unknown

Formula: Q = m x L_{ice}

Substitution: Q = 25600 x 336

Q = 8601600 J

(2) The quantity of heat that is required to change X g of ice to liquid is 235J. If the L_{ice} is 336J/g, calculate the mass of the object.

**Solution:**

Data given in the question:

Mass = Unknown,

Q = 235 J,

Lice = 336J/g

Formula: Q = m x Lice

Substitution: 235 = m x 336

Make m the subject: M = 235 ÷ 336

M = 0.699 g

Experiment to Determine the Latent Heat of Fusion of Ice

**Aim:**

To determine the latent heat of fusion of ice.

**Apparatus:**

Beaker, funnel, heater, ice block,

**Setup Diagram:**

**Procedures:**

Insert the heater into ice block contained in a funnel. Switch on the heater so that current flow for some times t. Adjust the rheostat so that steady current flow. Collect the melted ice in a beaker and measure its mass.

**Data:**

Current that flow in the heater = I

The voltage supplied = V

Time for which current flow = t

The resistance of the heater = R

Mass of melted ice = M_{ice}

Specific latent heat of fusion of ice = L

Theory of Calculation: –

Quantity of Heat Supplied by heater = Quantity of Heat Gained by melted ice

**The formula of Calculation:**

I²Rt = M x L_{ice}

**Precautions:**

- Use dry ice only
- Add a small quantity of ice at a time
- Stir the mixture gently and continuously for equal temperature.
- Lag the calorimeter to prevent heat loss.
- The amount of temperature below and above room temperature for the final and initial temperature should be the same.

**Effect of Expansion and Contraction on Fusion**

- Glass bottle of water crack on freezing because water expands on freezing and because of unequal expansion of the bottle.
- When water freezes into ice, it expands, become less dense and float on water with nine – tenth (9/10) of its volume submerged in water. Ice contracts on melting.

**Effect of Pressure on Freezing Point of a Substance**

- Increase in pressure lower the melting point of ice or the freezing point of water. This is called regelation of ice.
- An increase in pressure lowers the freezing point of any liquid which expands on solidifying. E.g. Water.
- Increase in pressure increases the freezing point of any substance which contract on solidifying. E.g. paraffin-wax.

**Definition of Regelation of Ice block**

Regelation of ice is the process whereby when ice block is subjected to high pressure, its melting point increases and the ice block melts, but when the pressure is removed, its freezing point increase and the ice refreezes.

**Effect of Impurities on Freezing Point of Substance**

The presence of impurities in a substance lowers the melting point of a pure substance

Specific latent heat of vaporization of steam is the quantity of heat energy that is required to change 1kg mass of a liquid into steam/gas at constant temperature ( boiling point) and pressure. It is a scalar quantity. The unit of specific latent heat of vaporization of steam is J/Kg or J/g.

Relationship between Specific Latent Heat of Fusion of Ice Lice and Specific Latent Heat of Vaporization Of Steam L_{steam}

Specific latent heat of fusion of ice Lice = 1/7 of specific latent heat of vaporization of steam

L_{ice} = 1/7 x L_{steam}

**Worked Examples:**

(1) If the latent heat of fusion of ice is 336J/g, what is e specific latent heat vaporization of steam?

**Solution**

Data given in the question:

Specific latent heat of fusion of ice Lice = 336j/g. L_{steam} = Unknown

**Formula:**

specific latent heavy of fusion of ice Lice = 1 /7 Specific latent heat of steam

Lice = 1/7 of L_{steam}

Substitution: 336 = 1/7 of L_{steam}

Make L_{steam} the subject: L_{steam} = 7 x 336

L_{steam} = 2352J/g

(2) What is the specific latent heat of fusion of ice if that of steam is 2352J/g?

Solution:

Data given in the question:

Specific latent heat of vaporization of steam L_{steam} = 2352J/g, Lice = Unknown

Formula: L_{ice} = 1/7 of L_{steam}

Substitution: L_{ice} = 1/7 x 2352.

L_{ice} = 336 J/g

In our next class, we will be talking about** the Gas Laws.** We hope you enjoyed the class.

Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.

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