5 gm of steam at `100^(@)C` is passed into six gm of ice at `0^(@)C`. If the latent heats of steam and ice in cal per gm are 540 and 80 respectively, then the final temperature is -

10 grams of steam at 100^(@)C is mixed with 50 gm of ice at 0^(@)C then final temperature is

10 grams of steam at 100^(@) C is mixed with 50 gm of ice at 0^(@) C then final temperture is

300 gm of water at 25^@ C is added to 100 gm of ice at 0^@ C. The final temperature of the mixture is

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

1 g of a steam at 100^(@)C melt how much ice at 0^(@)C ? (Length heat of ice = 80 cal//gm and latent heat of steam = 540 cal//gm )