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)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}