# Question #0624a

##### 1 Answer

Heat required**143.2Kcal**

#### Explanation:

The process can be split into 3 steps for simplification:

200 gms of ICE at

200 gms of WATER at

1 **. In first step**

200 gms of **ICE** at **WATER** at

The heat energy required is

[ **Quantity of heat required to change 1 Kg of a substance from solid state to liguid state at its Melting point}**

**Latent heat of fusion of ICE #=#80cal/gm**

So heat required for 200gm

2.) 200 gms of WATER at

**Q #=#mS#Delta#T** {m

SPECIFIC HEAT OF WATER

Q

3.) 200 gms of WATER at

Heat energy required is

** #color(blue)(LATENT. HEAT. OF. VAPORIZATION. OF. WATER)#**}

{ Quantity of heat energy required to change its 1 kg mass from

liquid to vapour state at its boiling point,}

Latent heat of vaporisation of water:

536Kcal/gm**540cal/gm**

So heat required for vaporisation of 200gm water to steam at

The amount of heat (in Joules) required to convert 200 grams of ice

to gas at 100 degrees celsius is the sum heat of 3 steps

**143.2Kcal**{144Kcal