Home
Class 11
PHYSICS
A refrigeratior converts 50 gram of wate...

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)`

Promotional Banner

Similar Questions

Explore conceptually related problems

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)

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 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 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 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 )

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) .

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)

Stream at 100^(@)C is passed into 20 g of water at 10^(@)C . When water acquires a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water =1 cal g^(-1) .^(@)C^(-1) and latent heat of steam =540 cal g^(-1) ]