How many grams of ice at ` -14 .^(@)C` are needed to cool 200 gram of water form `25 .^(@)C` to `10 .^(@)C`? Take specific heat of ice `=0.5 cal g^(-1) .^(@)C^(-1)` and latant heat of ice = `80 cal g^(-1)` .

How many gram of ice at -14^@C are needed to cool 200 gm of water from 25^@C to 10^@C . Specific heat of ice = 0.5 cal/gm ^@C and latent heat of ice = 80 cal/gm.

A refrigeratior converts 50 gram of water at 15^(@)C intoice at 0^(@)C in one hour. Calculate the quantity of heat removed per minute. Take specific heat of water =1 cal g^(-1) .^(@)C^(-1) and latent heat of ice =80 cal g^(-1)

10 gm of ice at – 20^(@)C is added to 10 gm of water at 50^(@)C . Specific heat of water = 1 cal//g–.^(@)C , specific heat of ice = 0.5 cal//gm-.^(@)C . Latent heat of ice = 80 cal/gm. Then resulting temperature is -

What amount of ice at -14^(@)C required to cool 200gg of water from 25^(@)C to 10CC? (Given C_("ice")=0.5cal/g.^(@)C,L_(f) for ice =80cal//g )

500 gms of water is to be cooled from 3.5^@C to 20^@C . How many grams of ice at -10^@C will be required for this purpose? Specific heat of ice = 0.5 cal g^(-1) ""^@C^(-1) Latent heat of ice = 80 cal g^(-1) ?

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)

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) ]