`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)`.

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,) .

50 g of ice at 0^(@)C is converted to steam at 100^(@)C . Calculate the amount of heat given in this process. (Latent heat of vapourisation= 540 cal. per gm)

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)

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.

Calculate the amount of heat required to convert 10g of water at 30^(@)C into steam at 100^(@)C . ( Specific latent heat of vaporization of water = 540 cal//g )

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 ]