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# Latent heat

Latent Heat

Latent heat or hidden heat is experienced when there is a change of state ( melting, vaporization, condensation, freezing, sublimation), it is not visible in the thermometer because there is no change in temperature. There are latent heat of fusion and latent heat of vaporization.

Latent heat of fusion is the heat energy required to convert a substance from its solid form to its liquid form without change in temperature.

Specific latent heat of fusion of a substance is the quantity of heat required to change unit mass of a substance from solid to liquid without change in temperature. The unit is Jkg-1.

Q = quantity of heat(in joules)

m = mass of substance(kg)

Dried ice is dropped in weighed calorimeter(M1) containing water of known mass(M2-M1) and known temperature(θ1). The mixture is stirred continuously and more ice is added until the temperature of the mixture falls to about 100C below the room temperature(θ2). The content is reweighed to find the mass of the ice.

Heat lost by calorimeter and water in cooling from θ1 to θ2 = Heat gained by ice in melting to water at 00c  + Heat gained by melted ice when its temperature rises from 00C to θ2.

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The ice must be dried before it is put in the calorimeter.

Example 3:

Calculate the heat energy required to   change 0.1kg of ice at  to water  boiling at . [Specific heat  capacity of water = 4200Jkg-1K-1;  specific latent heat of fusion of ice =336000Jkg-1].

Solution:

Q = Heat required to melt ice at 00C  +  Heat required to change temperature of ice from 00C to 1000C.

EVALUATION

1. Calculate the heat required to convert 20g of ice at to water at .  [Specific latent heat of fusion   of ice =  336Jg-1; specific heat capacity of water  = 4.2Jg-1K-1]
2. Explain what is meant by the following statements: The specific latent heat of fusion of ice is 3.4 x 105Jkg-1.

## Determination of Specific Latent Heat of Vaporization of Steam

Q = ml

l=Qm

l = specific latent heat of fusion

Specific latent heat of vaporization of a substance is the quantity of heat required to change unit mass of substance from liquid to vapour without change in temperature.

Q=mll=Qm

Example 1:

How much heat is required to convert 20g of ice at 0℃  to water at the same  temperature? (Specific latent heat of ice = 336Jg-1)

M=20gl=336Jg−1l=QmQ=mlQ=336×20=6720J

Example 2:

Calculate the quantity of heat released  when 100g of steam at 100℃ condenses to water. [Take specific latent heat of vaporization of water as Jkg-1]

M=100g=0.1kgl=2.3×106Jkg−1l=QmQ=mlQ=2.3×106×0.1×105J

EVALUATION

Calculate the energy required to vapourise 50g of water initially at 800C. [Specific heat capacity of  water = 4.2Jg-1K-1; specific latent  heat of vapourisation of water =  2260Jg-1]

## Determination of Specific Latent Heat of Fusion of Ice

The calorimeter is weighed empty and the mass(M1) is recorded. Water is poured into the calorimeter and the mass(M2) recorded. Dried steam is passed into the lagged calorimeter containing water until the temperature of water rises to 250and the steam is removed and the content stirred. The mass(M3 )is recorded and the final steady temperature taken(θ2).

Mass of water = M2 – M,   Mass  of steam = M3 – M2

Heat lost by stem in condensing  + Heat lost by condensed stem in cooling from 1000C to θ2  = Heat gained by water and calorimeter during the experiment.

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The Steam must be dried

Example 4:

Calculate the energy required to vaporize 50g of water initially at 800C . [Specific heat capacity of water = 4.2Jg-1K-1; specific latent  heat of vaporization of water = 2260Jg-1]

Solution:

Q = Heat required to raise the temperature of water from 800C to 1000C  +  Heat required to vaporize water

Example 5:

Heat is supplied to a liquid of mass 500g contained in a can by passing a current of 4A through a heating coil of resistance 12.5Ω immersed in the liquid. The initial temperature of the liquid is 240C. The liquid reaches its boiling point in 10 minutes after the current is switched on. It takes a further 2 minutes after the liquid starts to boil away. Calculate

(a). The specific heat capacity of the liquid

(b). The specific latent heat of vaporization of the liquid (boiling point of liquid = 840C, thermal capacity of can = 400J/K)

Solution:

(a) Mass of liquid = 500g = 0.5kg

Heat required to raise temperature of liquid from 240C to 840C (boiling point of liquid) is given as

Q=mc(θ2−θ1)=0.5×c×(84−24)=30c

c is the specific heat capacity of the liquid.

Heat required to raise temperature of can from 240C to 840C =400×60=24000J  (thermal capacity × change in temperature).

Heat supplied by heating coil is

IVt=I2Rt=4×4×12.5×10×60=120000J

Since this heat is used to raise the temperature of the can and the liquid to boiling point, we have

30c+24000=12000030c=120000–24000c=3200Jkg−1

(b) Let L be the specific latent heat of vaporization of the liquid.

Heat required to vaporize liquid =mL=0.5L

Heat supplied by current =I2Rt=4×4×12.5×2×60=24000J

Since this heat is required to boil away the liquid at 840C, we have

0.5L=24000 (neglecting heat loss to the surrounding)

L=240000.5=4.8×104Jk−1

GENERAL EVALUATION ( POST YOUR ANSWER IN THE COMMENT BOX AT THE BOTTOM  FOR RATING AND DISCUSSION)

1. A cup containing 100g of pure water at 20oC is placed in a refrigerator. If the refrigerator extracts heat at the rate of  840J per minute, calculate the time taken for the water to freeze. [Neglect  the heat capacity of the material of  the cup.] [Specific heat capacity of water = 4.2Jg-1K-1; specific latent heat of fusion of ice = 336Jg-1].