Latent Heat
Q1

3.5 kilograms of steam at 100°C is changed into water at 100°C. How much heat is given out ? (Latent heat of steam = 22.5 × 105 J/kg)

Sol:

Here steam (which is a gas) is changing into water (which is a liquid).
So, a change of state is taking place here. So, the formula to be applied is:

Q = m × L (for change of state)
Here, Mass of steam, m = 3.5 kg
And, Latent heat of steam, L = 22.5 × 105 J/kg
So, putting these values in the above formula, we get:
Q = 3.5 × 22.5 × 105
= 78.75 × 105 joules
Thus, the heat given out is 78.25 × 105 joules.
Q2

Calculate the amount of heat required to convert 620 g of ice into water without change of temperature? (Latent heat of ice = 3.34 × 105 J/kg)

Sol:

Here, solid ice is being converted into liquid water, so a change of state takes place.
The heat involved in the change of state is calculated by using the formula :

Q = m × L
Here, Mass of ice, m = 620 g
= 0.62 kg
And, Latent heat of ice, L = 3.34 × 105 J/kg
Now, putting these values in the above formula, we get:
Heat required, Q = 0.62 × 3.34 × 105
= 2.07 × 105 joules
Thus, the heat required is 2.07 × 105 joules.
Q3

Calculate the amount of heat required to convert 275 g of water at 100°C to steam at the same temperature? (Latent heat of vaporization of water = 22.5 × 105 J/kg)

Sol:

In this case, liquid water is being converted into a gas called steam, so a change of state takes place here.
The heat involved in the change of state of a substance is calculated by using the formula :

Q = m × L
Here, Mass of water, m = 275 g
= 0.275 kg
Latent heat of vaporization of water, L = 22.5 × 105 J/kg
Now, putting these values in the above formula, we get :
Q = 0.275 × 22.5 × 105
= 6.19 × 105 joules
Thus, the heat required is 6.19 × 105 joules.
Q4

288g of molten lead is at m.p. is just kept in molten state by a healer of 38 W. If the heater is switched off the temperature of lead starts falling after 3.4 min. Calculate the sp. latent beat of fusion of lead?

Sol:

Time for which temp, does not fall = Time in which lead changes its state.

Heat given out by lead at its m.p. = mL
= 288 g × L
Also, heat given out by lead = 38 Js−1 × 3.4 × 60s
= 7752 J
∴ 288 g × L = 7752 J
∴ L = 7752 J / 288 g
= 26.9 Jg−1.
Q5

A kilowatt immersion heater, is placed in 1.1 kg of crushed ice at 0°C and switched on for 5 min, when all the ice just melts. Calculate the sp. latent of fusion of ice ?

Sol:
Heat supplied by heater = 1100 J/s−1 × 5 × 60 s
= 330000 J
Heat required to melt ice = I kg × L
∴ Heat req. by ice = Heat supplied by heat
lkg × L = 330000 J
∴ L = 330000 J kg−1