Question #0624a

1 Answer
Dec 5, 2017

Heat required= 143.2Kcal

Explanation:

The process can be split into 3 steps for simplification:

200 gms of ICE at 0^0Ccolor(red)rArr200 gms of WATER at 0^0Ccolor(blue)rArr

200 gms of WATER at 100^0Ccolor(green)rArr200 gms of STEAM at 100^0C

1 . In first step

200 gms of ICE at 0^0Ccolor(red)rArr200 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 200gmrArr80(cal)/(gm)xx200gmrarr1600cal =16Kcal

2.) 200 gms of WATER at 0^0Ccolor(blue)rArr200 gms of WATER at 100^0

Q=mSDeltaT {mrarrmass:Srarrspecific heat; TrarrTemperature}

SPECIFIC HEAT OF WATERrarr1 cal/g-C^0

Q=200cancel(gm)xx1 cal/cancel(gm-C^0)xx(100-0)cancel(C^0)rArr20Kcal

3.) 200 gms of WATER at 100^0Ccolor(blue)rArr200 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/gmrarr536cal/gmhArr 540cal/gm

So heat required for vaporisation of 200gm water to steam at 100^0CrArr 536(cal)/cancel(gm)xx200cancel(gm)=107.2Kcal{108Kcalrarr 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

rArr16Kcal+20Kcal+107.2Kcal= 143.2Kcal{144KcalrarrIf you use L.H.V as 540cal/g}