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

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`= 2400 J kg ^(-1)K^(-1)` and the specific latent heat of ice `= 3.4 xx 10^(3)J kg^(-1)`

Text Solution

Verified by Experts

Heat released by 0.15 kg of water in being cooled to `0^@C=0.15xx1xx20=3kcal` Heat absorbed by ice from `-10^@`C to `0^@C=0.1xx0.5xx10=0.5kcal`
The balance heat is available for malting ice. Let m kg of ice melt.
Then `mxx80=2.5` or `m=0.03kg`
Thus the final temperature is `0^@C` with `0.07kg` of ice and `0.18kg` of water
Promotional Banner

Topper's Solved these Questions

  • CALORIMETRY

    CENGAGE PHYSICS|Exercise Exercise 1.1|23 Videos

  • CALORIMETRY

    CENGAGE PHYSICS|Exercise Exercise 1.2|22 Videos

  • BASIC MATHEMATICS

    CENGAGE PHYSICS|Exercise Exercise 2.6|20 Videos

  • CENTRE OF MASS

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos

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^(@) 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 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) ).

What will be the result of mixing 2kg of cooper at 100^(@)C with 0.75 kg of ice at 0^(@)C (Specific heat capacity of cooper =378J kg^(-1)K^(-1) specific heat capacity of water =4200 J kg^(-1)K^(-1) and specific latent heat of fusion of ice =336xx10^(3)Jkg^(-1) )

One kilogram of ice at 0^(@)C is mixed with one kilogram of water at 80^(@)C . The final temperature of the mixture is (Take : specific heat of water =4200 J kg^(-1) K^(-1) , latent heat of ice = 336 kJ//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) )

A lump of ice of 0.1 kg at -10^(@)C is put in 0.15 kg of water at 20^(@)C . How much water and ice will be found in the mixture whenit has reached thermal equilibrium? Specific heat of ice =0.5kcal kg^(-1).^(@)C^(-1) and its latent heat of melting =80 kcal kg^(-1)

Ice of mass 600 kg and at a temperature of -10^@C is placed in a copper vessel heated to 350^@C . The resultant mixture is 550 g of ice and water. Find the mass of the vessel. The specific heat capacity of copper (c)=100 cal//kg-K

A vessel containing 100 g ice at 0^(@)C is suspended in a room where temperature is 35^(@)C . It was found that the entire ice melted in 10 hour. Now the same vessel containing 100 g of water at 0^(@)C is suspended in the same room. How much time will it take for the temperature of water to rise to 0.5^(@)C . Neglect the heat capacity of the vessel. Specific heat of water and specific latent heat of fusion of ice are 1 cal g^(-1) .^(@)C^(-1) and 80 cal g^(-1) respectively.