Home
Class 10
PHYSICS
10 g of ice at 0°C absorbs 5460 J of hea...

10 g of ice at 0°C absorbs 5460 J of heat energy to melt and change to water at `50^@`C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J `kg^(-1) K^(-1)`.

Text Solution

AI Generated Solution

To calculate the specific latent heat of fusion of ice, we will follow these steps: ### Step 1: Convert the mass of ice from grams to kilograms The mass of ice is given as 10 g. To convert this to kilograms: \[ \text{mass} = \frac{10 \text{ g}}{1000} = 0.01 \text{ kg} \] ...
Promotional Banner

Topper's Solved these Questions

  • CALORIMETRY

    ICSE|Exercise EXERCISE-11(B) (MULTIPLE CHOICE TYPE)|2 Videos

  • WORK ENERGY AND POWER

    ICSE|Exercise EXERCISE-2(C ) NUMERICALS|6 Videos

  • CURRENT ELECTRICITY

    ICSE|Exercise EXERCISE-8(C) NUMERICALS |35 Videos

Similar Questions

Explore conceptually related problems

The specific latent heat of fusion of water is :

The amount of heat energy required to convert 1 kg of ice at -10^@ C to water at 100^@ C is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg^(-1) K^(-1) , specific heat capacity of water = 4200 J kg^(-1) K^(-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) .

A metal piece of mass 20 g is heated to a constant temperature of 100^@ C. Then it is added in a calorimeter of mass 50 g and specific heat capacity 0.42 J g^(-1) K^(-1) , containing 50 g of water at 20^@ C. After stirring the water, the highest temperature recorded is 22^@ C. Calculate the specific heat capacity of metal. Specific heat capacity of water = 4.2 J g^(-1)K^(-1)

40 g of ice at 0^(@)C is used to bring down the temperature of a certain mass of water at 60^(@)C to 10^(@)С . Find the mass of water used. [Specific heat capacity of water = 4200 Jkg^(-1)""^(@)C^(-1) ] [Specific latent heat of fusion of ice = 336xx10^(3)Jkg^(-1) ]

A vessel of mass 100 g contains 150 g of water at 30^@ C. How much ice is needed to cool it to 5^@ C ? Take specific heat capacity of material of vessel = 0-4 J g^(-1) K^(-1) , specific latent heat of fusion of ice = 336 J g- and specific heat capacity of water = 4.2 J g^(-1) K^(-1) .

200 g of hot water at 80^@ C is added to 300 g of cold water at 10^@ C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg^(-1)K^(-1)

Calculate the heat energy required to raise the temperature of 2 kg of water from 10^@ C to 50^@ C. Specific heat capacity of water is 4200 J kg^(-1) K^(-1) .

250 g of water at 30^(@)C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5^(@)C . Specific latent heat of fusion of ice = 336xx10^(3)Jkg^(-1) Specific heat capacity of copper vessel = 400Jkg^(-1)""^(@)C^(-1) Specific heat capacity of water = 4200Jkg^(-1)""^(@)C^(-1) .