`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`.

50 g of steam at 100^@ C is passed into 250 g of at 0^@ C . Find the resultant temperature (if latent heat of steam is 540 cal//g , latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g -.^@ C ).

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 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 )

75 g of water at 100^@ C is added to 20 g of ice at -15^@C . Calculate the resulting temperature, give that latent heat of ice= 80 cal g^(-1)C^(-1) and specific heat of ice=0.5 cal g^(-1) C^(-1) .