A refrigerator converts `100` g of water at `25^(@)C` into ice at `-10^(@)C` in one hour and `50` minutes. The quantity of heat removed per minute is (specific heat of ice = `0.5 "cal"//g^(@)C`, latent heat of fusion = `80 "cal"//g`)

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)

100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and 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)

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 -

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

A piece of ice of mass 50 g exists at a temperature of -20^(@)C . Determine the total heart required to covert it completely to steam at 100^(@)C . (Specific heat capacity of ice = 0.5 cal//g-^(@)C , specific latent heat of fusion for ice = 80 cal//g and specific latent heat of vaporization for water = 540 cal//g ).

Find the heat required to convert 1 g of ice at 0°0C to vapour at 100°C. (Latent heat of ice = 80 cal/g and latent heat of vapour = 536 cal/g)