Melting, Boiling and Evaporation DEFINITIONS Melting point: it is the temperature at which a solid changes into a liquid or vice versa. Boiling point: it is the temperature at which a liquid changes into a gas or vice versa. Heat capacity (C): it is the amount of heat energy needed to change the temperature of a substance by 1 Kelvin (or 1oC).

Specific heat capacity (c): it is the amount heat energy needed to change the temperature of 1 kg of a substance by 1K (or 1oC).

Latent heat: it is the amount of heat energy needed to change the state of an object. Specific latent heat (L): it is the amount of heat energy needed to change the state of 1 kg of an object. Specific latent heat of fusion (Lf): it is the amount of heat energy needed to change 1 kg of a solid into liquid.

Specific latent heat of vaporization (Lv): it is the amount of heat energy needed to change 1 kg of a liquid into gas.

HEATING CURVE Temperature Boiling Point

Gas liquid + gas

Liquid Melting point

solid + liquid

Solid Time As a solid is heated its temperature increases until it reaches the melting point. The amount of heat energy required for this temperature rise can be calculated using the formula: = â&#x2C6;&#x2020; . At the melting point the solid starts to melt. There is no temperature change during melting because all the heat energy given to the solid is used to break the bonds and separate the molecules. Temperature is a measure of the kinetic energy of an object. As there is no change in the kinetic energy of the solid there is no change in its temperature during melting. The amount of heat energy required to melt the solid at this temperature can be calculated using the formula: = . When the entire solid has changed into a liquid its temperature starts rising again until it reaches the boiling point. The temperature again does not change during boiling as all the heat energy provided at this point is used to break the bonds and increase spaces between molecules of the liquid. Once the liquid fully changes into gas its temperature keeps on rising further.

CHANGES OF STATE

SOLID SOLID

Melting

solidification

LIQUID

GAS

Boiling

LIQUID

Condensation

GAS

EVAPORATION It is a process in which the liquid gains heat energy from the surroundings and the more energetic molecules escape from the surface of the liquid. The average kinetic energy of the remaining molecules drops and so the temperature of the liquid also drops. This is why evaporation causes cooling. For example: if in a class there are students with 97%, 95%, 98%, 54%, 43%, and 73 %. The average of these students will be 77%. If the tree students with 97%, 95% and 98% leave the class then the average will become only 57%. Similarly if the more energetic molecules leave the liquid then the average kinetic energy of the liquid will drop and its temperature will also decrease as temperature is a measure of average kinetic energy.

DIFFERENCES BETWEEN EVAPORATION AND BOILING

1. 2. 3. 4.

Evaporation It takes place at any temperature It is a slow process It takes place only on the surface of the liquid No bubbles are formed

1. 2. 3. 4.

Boiling It takes place at a fixed temperature It is a fast process It takes place throughout the liquid Bubbles are formed

FACTORS AFFECTING EVAPORATION 1. Temperature: the higher the temperature the greater is the rate of evaporation. 2. Humidity: rate of evaporation decreases with humidity. The higher the humidity the lower is the rate of evaporation. 3. Wind: wind makes the evaporated molecules to fly away from top of the liquidâ&#x20AC;&#x2122;s surface thus creating space for more molecules to evaporate. So wind increases the rate of evaporation. 4. Air pressure: if outside air pressure is less it is easier for the molecules to escape out of the liquidâ&#x20AC;&#x2122;s surface. So low air pressure on the outside increases the rate of evaporation. This is why liquid boils at a lower temperature on top of mountains where there is less air pressure outside. 5. Surface area: if the surface area is more then there is more space for the molecules to evaporate into the air. So a larger surface area increases the rate of evaporation.

NOTE:

depth of liquid does not have any effect on the rate of evaporation.

Heat capacity (C): it is the amount of heat energy needed to change the temperature of a substance by 1 Kelvin (or 1oC). Formula:

=

Or

= ∆

unit of heat capacity: J/K

Where C = heat capacity (J/K) Q = Amount of heat energy (J) ∆θ = change in temperature (K)

Specific heat capacity (c): it is the amount heat energy needed to change the temperature of 1 kg of a substance by 1K (or 1oC). Formula:

=

Or

=

unit of specific heat capacity: J/kgK

Where c = specific heat capacity (J/kgK) Q = Amount of heat energy (J) ∆θ = change in temperature (K)

Specific latent heat (L): it is the amount of heat energy needed to change the state of 1 kg of an object. Formula:

=

Or

=

unit of latent heat: J/kg

Where L = latent heat (J/kg) Q = Amount of heat energy (J) m = mass of the substance (kg)

REMEMBER:

Always use indicated, such as ice melting at 0oC.

=

when there is a change of state and no temperature change is

Always use = ∆ when there is a temperature change such as a liquid heating from 10 C to 50 C. This formula cannot be used when there is a change of state. o

Example:

o

Calculate the total amount of heat energy needed to change 2 kg ice at -10oC to water at 30oC. Specific heat capacity of ice = 2100 J/kgK Specific heat capacity of water = 4200 J/kgK Specific latent heat of fusion of ice = 335000 J/kg

firstly calculate the energy needed to raise the temperature of 2 kg of ice from -10oC to 0oC.

Solution:

Data: mass of ice, m = 2 kg; c = 2100 J/kgK;

change in temperature, ∆ = 10oC.

Q1 = mc∆ = 2 x 2100 x 10 = 42000 J Now calculate the energy needed to melt 2 kg of ice at 0oC. Data: mass of ice, m = 2 kg; L= 335000 J/kg ;

Q2 = mL = 2 x 335000 = 670 000 J Now you have to calculate the heat energy needed to raise 2 kg of water (melted ice) from 0oC to 30oC. change in temperature, ∆ = 30oC.

Data: mass of water, m = 2 kg; c = 4200 J/kgK ;

Q3 = mc∆ = 2 x 4200 x 30 = 252000 J Total heat energy = 42000 + 670000 + 252000 = 964 000 J.

Experiment:

To find the find the specific heat capacity of aluminium.

Apparatus: an aluminium block, electric heater, thermometer, stopwatch, ammeter, voltmeter, cotton. Diagram:

thermometer A

Method: 1.

V

2. 3. 4. 5.

Formula:

cotton lagging Where V = voltage (V);

Record the mass of the aluminium block and its initial temperature. Start the heater and the stopwatch simultaneously. Take the ammeter and voltmeter readings. Record the final temperature of the block. Use the formula given below to calculate the specific heat capacity of the aluminium.

I = current (A);

∆ = change in temperature (oC);

t = time (s);

=

m =mass of the aluminium block (kg); c = specific heat capacity of aluminium (J/kg/K)

Note: you can find the specific heat capacity of a liquid in a similar way. Only put the liquid in a beaker and the rest of the method is same. Precautions:

1.

use lagging to minimise heat loss to surroundings.

2.

Wait until the mercury thread in the thermometer stops rising and then take the maximum reading of the thermometer.

Thermal Properties of Matter

Boiling Point liquid + gas When the entire solid has changed into a liquid its temperature starts rising again until it reaches the boiling...