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The final temperature, when 10 g of stea...

The final temperature, when `10 g` of steam at `100^@ C` is passed into an ice block of mass `100 g`
`(L_(steam) = 540 cal//g, L_(ice) = 80 cal//g , S_(water) =1 cal//g^(@) C)` is

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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) .

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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 )

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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 )

1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )

1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//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,) .