1 gm of ice at 0^@C is converted to stea...

`1 gm` of ice at `0^@C` is converted to steam at `100^@C` the amount of heat required will be `(L_("steam") = 536 cal//g)`.

100 calorie

`0.01` kilocalorie

716 calorie

1 kilocalorie

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1 gram of ice at -10^@ C is converted to steam at 100^@ C the amount of heat required is (S_(ice) = 0.5 cal//g -^(@) C) (L_(v) = 536 cal//g & L_(f) = 80 cal//g,) .

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