A iron ball of mass `0.2 kg ` is heated to `100^(@)C` and put into a block of ice at `0^(@)C`.25 g of ice melts , then specific heat of iron ( in cal` kg^(-10) C^(-1) )` is [ Latent heat of fusion of ice = `80 cal g^(-1)`]

An iron ball of mass 0.2 kg is heated to 10^(@)C and put into a block of ice at 0^(@)C . 25 g of ice melts. If the latent heat of fusion of ice is 80 cal g^(-1) , then the specific heat of iron in cal g^(-1^(@))C is

An iron ball of mass 0.2 kg is heated to 10^(@)C and put into a block of ice at 0^(@)C . 25 g of ice melts. If the latent heat of fusion of ice is 80 cal g^(-1) , then the specific heat of iron in cal g^(-1^(@))C is

10g of water at 30^(@)C and 5g of ice at -20^(@) C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice= 0.5 cal g^(-1) (""^(0)C)^(-1) and latent heat of fusion of ice =80 cal g^(-1)

Calculate the entropy change in the melting of 1 kg of ice at 0^(@)C in SI units. Heat of funsion of ice = 80 cal g^(-1) .

Calculate the entropy change in the melting of 1 kg of ice at 0^(@)C in SI units. Heat of funsion of ice = 80 cal g^(-1) .

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

10 gram of water at 45^(@)C is added to 5 gram of ice at -30^(@)C . Find the final temperature of the mixture. Specific heat of ice is 0.55 cal g^(-1)""^(@)C^(-1) and that of water is 1.0 cal g^(-1)""^(@)C^(-1) latent heat of fusion of ice is 80 cal g^(-1) ? (ii) To rise the temperature of 100 gram of water from 24^(@) to 90^(@)C by steam at 100^(@)C . Calculate the amount of steam required. Specific heat of water is 1.0 cal g^(-1)""^(@)C^(-1) . Latent heat of steam is 540 Cal kg^(-1) .