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**3. Phase change : **

Heat required for the change of phase or state,

**Q = mL, **L = latent heat.

**(a)** **Latent heat (L) : **The heat supplied to a substance which changes its state at constant temperature is called latent heat of the body.

**(b)** **Latent heat of Fusion (L _{f}) : **The heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm. pressure is called latent heat of fusion.

**(c)** **Latent heat of vaporisation (L _{v}) : **The heat supplied to a substance which changes it from liquid to vapour state at its boiling point and 1 atm. pressure is called latent heat of vaporization.

If in question latent heat of water are not mentioned and to solve the problem it require to assume that we should consider following values.

**Latent heat of ice : **L = 80 cal/gm = 80 Kcal/kg = 4200 Ã— 80 J/kg** **

**Latent heat of steam : **L = 540 cal/gm = 540 Kcal/kg = 4200 Ã— 540 J/kg

The given figure, represents the change of state by different lines

**Note :** If we increases the temperature of liquid (phase) Â but at a later time K .E. stop increasing and the phase of the liquid starts changing.

*Ex.5 Find amount of heat released if 100 g ice at - 10Â°C is converted into 120Â°C, 100 g steam.*

Q_{net} = 73.5 Kcal.

*Ex.6 500 gm of water at 80Â°C is mixed with 100 gm steam at 120Â°C. Find out the final mixture.*

**Sol. **

120Â°C steam 100Â°C steam

Req. heat = 100 Ã— Ã—20 = 1 kcal

80Â°C water 100Â°C water

Req. heat = 500 Ã— 1 Ã— 20 = 10 kcal

100gm steam 100 gm water at 100Â°C

Req. heat = 100 Ã— 540 = 54 kcal

Total heat = 55 kcal.

Remaining heat = 55 - 10 = 45 kcal

Now we have 600 gm water at 100Â°C â‡’ 4500 = m Ã— 540 â‡’

So at last we have gm steam and of water

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