# Specific Latent Heat

Card 1:The Heating Curve

## Heating Curve State of matter:
A-B: Solid
B-C: Solid and Liquid
C-D: Liquid
D-E: Liquid and Gas
E-F: Gas

Card 2: The Heating Curve - Latent Heat

## Heating Curve - Latent Heat • T1 is the melting point whereas T2 is the boiling point.
• From Q to R and S to T, the temperature remains constant because the heat supplied to the object is used to overcome the forces of attraction that hold the particles together.
• Heat obsorbs during Q-R is called the latent heat of fusion.
• Heat obsorbs during S-T is called the latent heat of vaporisation.
Card 3: The Cooling Curve

## Cooling Curve States of matter:
P-Q: Gas
Q-R: Gas and Liquid
R-S: Liquid
S-T: Liquid and Solid
T-U: Solid
U-V: Solid

Card 4: The Cooling Curve - Latent Heat

## Cooling Curve - Latent Heat • T1 is the condensation point, T2 is the freezing point whereas T3 is room temperature.
• During Q-R and S-T, the temperature remains unchanged. This is because the energy produced during the formation of bonds is equal to the heat energy released to the surroundings during cooling.
• The heat energy released during Q-R is called the latent heat of vaporization.
• The heat energy released during S-T is called the latent heat of fusion.
Card 5: What is Latent Heat?

## Latent Heat

Latent heat is the heat absorbed or releases during a change of state of matter.

Card 6: Define the specific latent heat of vaporisation

## Latent Heat of Vaporisation

The specific latent heat of vaporization is the heat needed to change 1 kg of a liquid at its boiling point into vapour or vice versa, without a change in temperature.

Card 7: Define specific latent heat of fusion.

## Latent Heat of Fusion

The specific latent heat of fusion is the heat needed to change 1 kg of a solid at its melting point into a liquid, or vice versa, without a change in temperature.

Card 8: Formula of Specific Latent Heat

## Formula o Specific Latent Heat Example 1
How much heat energy is required to change 0.5 kg of ice at 0°C into water at 25°C? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K).]

There are 2 processes involve when an ice is converted into water at 25°C.

Ice at 0°C ------> Water at 0°C -------> Water at 25°C

Energy absorbed to convert 0.5kg from Ice at 0°C to Water at 0°C

Q1 = mL = (0.5)(334000) = 167000J

Energy absorbed to convert 0.5kg from watyer at 0°C to Water at 25°C

Q2 = mcθ = (0.5)(4200)(25-0) = 52500J

Total energy required = Q1 + Q2 = 167000 + 52500 = 219500J

Example 2
How much energy is required to change exactly 1 g of ice at -20°C to steam at 120 °C? [Specific heat capacity of water = 4200J kg-1 oC-1; Specific latent heat of fusion of ice = 334,000 Jkg-1, specific heat capacity of steam is 2020 J/(kg °C), Specific latent heat of vaporization of water = 2,260,000 J/kg, specific heat capacity of ice = 2100 J/(kg K)]

The processes involved:

Ice (-20°C) -----> Ice (0°C) -----> Water (0°C) -----> Water (100°C)
-----> Steam (100°C) -----> Water (120°C)

Energy required:
Ice (-20°C) to Ice (0°C), Q1 = mcθ = (0.001)(2100)(20) = 42J
Ice (0°C) to Water (0°C), Q2 = mL = (0.001)(334000) = 334J
Water (0°C) to Water (100°C), Q3 = mcθ = (0.001)(4200)(100) = 420J
Water (100°C) to Steam (100°C), Q4 = mL = (0.001)(2260000) = 2260J
Steam (100°C) to Steam (120°C), Q5 = mcθ = (0.001)(2020)(20) = 40.4J

Total energy required
= Q1 + Q2 + Q3 + Q4 + Q5
= 42 + 334 + 420 + 2260 + 40.4 = 3096.4J

Card 9: What will happen to the melting point of ice if the ice contains impurities?

## How impurities affect the melting point

If impurities are present in a substance, its melting point will be lower than normal.

Card 10: Factors affecting melting point

## Factors affecting melting point

Pressure:
Applying pressure to ice, for example, lowers its melting point.

Presence of impurities:
Adding salt to melting ice, for example, can reduce its melting point to as low as -18°C. Copyright © 2008 Home Education Concept. All Rights Reserved 