Question

Question #0624a

Answer 1

Heat required`=` 143.2Kcal

Explanation:

The process can be split into 3 steps for simplification:

200 gms of ICE at `0^0Ccolor(red)rArr`200 gms of WATER at `0^0Ccolor(blue)rArr`

200 gms of WATER at `100^0Ccolor(green)rArr`200 gms of STEAM at `100^0C`

1 . In first step

200 gms of ICE at `0^0Ccolor(red)rArr`200 gms of WATER at `0^0C`

The heat energy required is

`color(red)(LATENT. HEAT .OF. FUSION. OF. ICE)`

[ 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`rArr`80`(cal)/(gm)` `xx`200gm`rarr`1600cal `=`16Kcal

2.) 200 gms of WATER at `0^0Ccolor(blue)rArr`200 gms of WATER at `100^0`

Q`=`mS`Delta`T {m`rarr`mass:S`rarr`specific heat; T`rarr`Temperature}

SPECIFIC HEAT OF WATER`rarr`1 cal/g-`C^0`

Q`=`200`cancel(gm)` `xx`1 cal/`cancel(gm-C^0)` `xx`(100-0)`cancel(C^0)` `rArr`20Kcal

3.) 200 gms of WATER at `100^0Ccolor(blue)rArr`200 gms of STEAM at `100^0C`

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`rarr`536cal/gm`hArr` 540cal/gm

So heat required for vaporisation of 200gm water to steam at `100^0C` `rArr` 536`(cal)/cancel(gm)xx`200`cancel(gm)` `=`107.2Kcal{108Kcal`rarr` if you use 540cal/gm}

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

`rArr`16Kcal+20Kcal+107.2Kcal`=` 143.2Kcal{144Kcal`rarr`If you use L.H.V as 540cal/g}