`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,)`.

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) .

10 g ice at 0^@C is converted into steam at 100^@C . Find total heat required . (L_f = 80 cal//g, S_w = 1cal//g-^@C, l_v = 540 cal//g)

0 g ice at 0^@C is converted into steam at 100^@C . Find total heat required . (L_f = 80 cal//g, S_w = 1cal//g-^@C, l_v = 540 cal//g)

5 g of steam at 100^(@)C is mixed with 10 g of ice at 0^(@)C . Choose correct alternative /s) :- (Given s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g )

5 g of steam at 100^(@)C is mixed with 10 g of ice at 0^(@)C . Choose correct alternative /s) :- (Given s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g )

5 g of steam at 100^(@)C is mixed with 10 g of ice at 0^(@)C . Choose correct alternative /s) :- (Given s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g )

How much heat is required to convert 8.0 g of ice at -15^@C to steam at 100^@C ? (Given, c_(ice) = 0.53 cal//g.^@C, L_f = 80 cal//g and L_v = 539 cal//g, and c_(water) = 1 cal//g.^@C) .

Ten grams of ice at - 20^(@)C is to be changed to steam at 130^(@)C . The specific heat of both ice and steam is 0.5 cal // ( g^(@)C) . The specific heat of water is 1.00 cal // ( g K ) . The heat of fusion is 80 cal // g and the heat of vaporization is 540 cal // g. The entire process requires.

6 gm of steam at 100^@ C is mixed with 6 gm of ice at 0^@ C . Find the mass of steam left uncondensed. (L_(f)=80 cal//g,L_(v)= 540cal//g, S_(water) = 1cal//g-^(@) C) .