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How much heat energy is released when 5 g of water at `20^(@)C` changes to ice at `0^(@)C`?
[Specific heat capacity of water = `4.2Jg^(-1)""^(@)C^(-1)`
Specific latent heat of fusion of ice = `336Jg^(-1)`]

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To find the total heat energy released when 5 g of water at 20°C changes to ice at 0°C, we need to consider two processes: 1. Cooling the water from 20°C to 0°C. 2. Freezing the water at 0°C to ice. ### Step 1: Calculate the heat released when cooling the water from 20°C to 0°C. The formula to calculate the heat released (Q1) when cooling is given by: ...
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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) .

The specific latent heat of fusion of water is :

104 g of water at 30^@ C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10^@ C. Find the mass of ice added. [Specific heat capacity of water = 4.2 J g^(-1)""^@ C^(-1) , Specific latent heat of fusion of ice = 336 J g^(-1) , Specific heat capacity of copper = 0.4 Jg^(-1) ""^@ C^(-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) ]

How much heat energy is gained when 5 kg of water at 20^(@)C is brought to its boiling point (Specific heat of water = 4.2 kj kg c)

A refrigerator converts 100 g of water at 20^(@)C to ice at -10^(@)C in 35 minutes. Calculate the average rate of heat extraction in terms of watts. Given : Specific heat capacity of ice = 2.1Jg^(-1)""^(@)C^(-1) Specific heat capacity of water = 4.2Jg^(-1)""^(@)C^(-1) Specific Latent heat of fusion of ice = 336Jg^(-1)

A piece of ice of mass 60 g is dropped into 140 g of water at 50^(@)C . Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding) Specific heat capacity of water = 4.2Jg^(-1)k^(-1) Specific latent heat of fusion of ice = 336Jg^(-1)

Heat energy is supplied at a constant rate to 100 g of ice at 0^(@)C . The ice is converted into water at 0^(@)C in 2 minutes. How much time will be required to raise the temperature of water from 0^(@)C" to "20^(@)C ? [Given : sp. Heat capacity of water 4.2g^(-1)""^(@)C^(-1) , sp. latent heat of ice = 336Jg^(-1) ]

What will be the result of mixing 400 g of copper chips at 500^@ C with 500 g of crushed ice at 0^@ C ? Specific heat capacity of copper = 0.42 J g^(-1) K^(-1) , specific latent heat of fusion of ice = 340 J g^(-1) .

200 g of water at 50.5^@C is cooled down to 10^@C by adding m g of ice cubes at 0^@C in it. Find m. Take, specific heat capacity of water = 4.2Jg^(-1)""^@C^(-1) and specific latent heat of ice = 336Jg^(-1)

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