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When m gram of steam at 100^(@) C is mix...

When m gram of steam at `100^(@) C` is mixed with 200 gm of ice at `0^(@)C`. it results in water at `40^(@) C`. Find the value of m in gram . (given : Latent heat of fusion (`L_f`) = 80 cal/gm, Latent heat of vaporisation (`L_v`) = 540 cal/gm., specific heat of water (`C_w`)= 1 cal/gm/C)

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A thermally insulated pot has 200 g ice at temperature 0^(@)C . How much steam of 100^(@)C has to be mixed to it, so that water of temperature 40^(@)C will be obtained? (Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporisation of water = 540 cal/g, specific heat of water = I " cal/g " ^(@)C)

4 gm of steam at 100^(@)C is added to 20 gm of water at 46^(@)C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal//gm . Specific heat of water =1 cal//gm-^(@)C .

1 kg ice at -10^(@)C is mixed with 0.1 kg of steam at 200^(@)C . If final temperature of mixture at equilibrium is T_(eq)=(58x)/11 , then fill the vallue of x Latent heat of fusion of ice =80 cal/gram, latent heat of vaporization of water =540 cal/gram, specific heat capacity of ice ~= specific heat of water =0.5 cal/gram-K

20 g of steam at 100^@ C is passes into 100 g of ice at 0^@C . Find the resultant temperature if latent heat if steam is 540 cal//g ,latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g^@C .

How many calories are required to change one gram of 0^(@)C ice to 100^(@)C steam ? The latent heat of fusion is 80 cal // g and the latent heat of vaporization is 540 cal // g. The specific heat of water is 1.00 cal // (g K ) .

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 )

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