`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)`

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) .

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) .

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) .

How much heat is required to convert 8.0 g of ice at -15^@ to steam at 100^@ ? (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) .

How much heat is required to convert 8.0 g of ice at -15^@ to steam at 100^@ ? (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) .

If there is no heat loss, the heat released by the condensation of x gram of steam at 100^@C into water at 100^@C can be used to convert y gram of ice at 0^@C into water at 100^@C . Then the ratio of y:x is nearly [Given L_l = 80 cal//gm and L_v= 540 cal//gm ]

If there is no heat loss, the heat released by the condensation of x gram of steam at 100^@C into water at 100^@C can be used to convert y gram of ice at 0^@C into water at 100^@C . Then the ratio of y:x is nearly [Given L_l = 80 cal//gm and L_v= 540 cal//gm ]

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 )