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The amount of heat required to convert 1gm of ice at 0 degree centigrade into steam at 100 degrees centigrade is .......[L ice=80,Sw=1,L steam=540]



It's would take 80 cal for converting I've to water, then 100 cal for bringing water from 1 to 100 degree & finally 540 cal for converting water at 100 degree to stream. Total 720 cal needed for 1g of ice.
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kk.i know that but i am unable to understand the procedure of the ans
And steam in place of stream in line 2, stupid auto correct;)
First, ice will convert into water using latent heat of ice which we need to supply. Then we need to bring water to its boiling point, i.e 100 degree for making it steam, finally we need to provide the latent heat of steam to make it steam at 100 degree temp.
you must consider your reference for further theory info
genius answer
The Brainliest Answer!
The amount of required to raise the temperature of a unit mass of substance through a unit rise in temperature, without the phase-change, is called the "Specific-Heat" of the substance.
s=\frac{\Delta Q}{m\Delta t}.
Whereas "Latent-Heat" is the amount of heat required to change the phase of a unit mass of substance.
L=\frac{\Delta Q}{m}.
Now for the phase change from solid-ice to liquid-water, heat required is
(1 gm)(80 cal/gm) = 80 cal.
To raise the temperature of water from 0 to 100 (celsius), heat required is
(1 gm)(1 cal/gm/K)(100 K) = 100 cal.
Again for the phase change from liquid-water to gaseous-steam, heat required is 
(1 gm)(540 cal/gm) = 540 cal.
Summing up the entire thing you get, total heat required = 720 cal = 0.72 kcal.
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