The water of mass 75 g at `100^(@)C` is added to ice of mass `20 g` at `-15^(@)C` . What is the resulting temperature. Latent heat of ice = `80 cal//g` and specific heat of ice = `0.5`.

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) .

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 .

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 .

One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -