[" 1gram of ice at "-10^(@)C" is converted to steam "],[" at "100^(@)C" the amount of heat required is "],[(S_(ice)=0.5cal/g-^(@)C)]

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) .

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.

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)

Work done in converting one gram of ice at -10^@C into steam at 100^@C is

The following graph represents change of state of 1 gram of ice at -20^@C . Find the net heat required to convert ice into steam at 100^@ C . S_(ice) = 0.53 cal//g -^@ C .

What is the amount of heat required (in calories) to convert 10 g of ice at -10^(@)C into steam at 100^(@)C ? Given that latent heat of vaporization of water is "540 cal g"^(-1) , latent heat of fusion of ice is "80 cal g"^(-1) , the specific heat capacity of water and ice are "1 cal g"^(-1).^(@)C^(-1) and "0.5 cal g"^(-1).^(@)C^(-1) respectively.