Home
Class 11
PHYSICS
A piece of ice of mass 100 g and at temp...

A piece of ice of mass `100 g` and at temperature `0^@ C` is put in `200 g` of water of `25^@ C`. How much ice will melt as the temperature of the water reaches `0^@ C` ? (specific heat capacity of water `=4200 J kg^(-1) K^(-1)` and latent heat of fusion of ice `= 3.4 xx 10^(5) J Kg^(-1)`).

Promotional Banner

Similar Questions

Explore conceptually related problems

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@)C .How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the specific latent heat of ice = 3.4 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass 40 g is added to 200 g of water at 50^@ C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg^(-1) K^(-1) and specific latent heat of fusion of ice = 336 xx 10^3" J "kg^(-1) .

If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

How much heat energy is released when 5.0 g of water at 20^@ C changes into ice at 0^@ C ? Take specific heat capacity of water = 4.2 J g^(-1) K^(-1) , specific latent heat of fusion of ice = 336 J g^(-1) .