**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(M_{1}) containing water of known mass(M_{2}-M_{1}) and known temperature(θ_{1}). The mixture is stirred continuously and more ice is added until the temperature of the mixture falls to about 10^{0}C below the room temperature(θ_{2}). The content is reweighed to find the mass of the ice.

Heat lost by calorimeter and water in cooling from θ_{1} to θ_{2} = Heat gained by ice in melting to water at 0^{0}c + Heat gained by melted ice when its temperature rises from 0^{0}C to θ_{2.}

M1CI(θ1−θ2)+(M2−M1)CW(θ1−θ)=(M3−M2)L+(M3−M2)CWθ1L=M1CI(θ1−θ2)+(M2−M1)CW(θ1−θ)−(M3−M2)CWθ1(M3−M2) |

**Precautions:**

- The calorimeter should be well lagged.
- The mixture should be well stirred to ensure even distribution of heat.
- 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^{-1}K^{-1}; specific latent heat of fusion of ice =336000Jkg^{-1}].

**Solution:**

Q = Heat required to melt ice at 0^{0}C + Heat required to change temperature of ice from 0^{0}C to 100^{0}C.

Q=ml+mcw(θ2−θ1)Q=0.1×336000+0.1×4200×(100−0)Q=33600+42000=75600=7.56×104J |

**EVALUATION**

- 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^{-1}K^{-1}] - Explain what is meant by the following statements: The specific latent heat of fusion of ice is 3.4 x 10
^{5}Jkg^{-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 80^{0}C. [Specific heat capacity of water = 4.2Jg^{-1}K^{-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(M_{1}) is recorded. Water is poured into the calorimeter and the mass(M_{2}) recorded. Dried steam is passed into the lagged calorimeter containing water until the temperature of water rises to 25^{0}and the steam is removed and the content stirred. The mass(M_{3} )is recorded and the final steady temperature taken(θ_{2}).

Mass of water = M_{2} – M_{1 }, Mass of steam = M_{3} – M_{2}

Heat lost by stem in condensing + Heat lost by condensed stem in cooling from 100^{0}C to θ_{2} = Heat gained by water and calorimeter during the experiment.

(M3−M2)L+(M3−M2)(100−θ2)CW=(M2−M1)CW(θ2−θ1)+M1C(θ2−θ1)L=(M2−M1)CW(θ2−θ1)+M1C(θ2−θ1)−(M3−M2)(100−θ2)CW(M3−M2) |

**Precautions:**

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

**Example 4:**

Calculate the energy required to vaporize 50g of water initially at 80^{0}C . [Specific heat capacity of water = 4.2Jg^{-1}K^{-1}; specific latent heat of vaporization of water = 2260Jg^{-1}]

**Solution:**

Q = Heat required to raise the temperature of water from 80^{0}C to 100^{0}C + Heat required to vaporize water

Q=mcw(θ2−θ1)+mlQ=50×42×(100−80)+50×2260Q=42000+113000=155000=1.55×105J |

**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 24^{0}C. 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 = 84^{0}C, thermal capacity of can = 400J/K)

**Solution:**

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

Heat required to raise temperature of liquid from 24^{0}C to 84^{0}C (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 24^{0}C to 84^{0}C =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 84^{0}C, 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)**

- A cup containing 100g of pure water at 20
^{o}C 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^{-1}K^{-1}; specific latent heat of fusion of ice = 336Jg^{-1}].

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