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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)`]

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To solve the problem step by step, we need to determine the heat energy supplied to convert ice into water and then calculate the time required to raise the temperature of the water from 0°C to 20°C. ### Step 1: Calculate the heat energy supplied to convert ice to water The heat energy required to convert ice at 0°C to water at 0°C can be calculated using the formula: \[ Q_1 = m \times L \] where: - \( m = 100 \, \text{g} \) (mass of ice), - \( L = 336 \, \text{J/g} \) (specific latent heat of ice). ...
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Heat energy is supplied at a constant rate to 400 g of ice at 0^@ C. The ice is converted into water at 0^@ C in 5 minutes. How much time will be required to raise the temperature of water from 0^@ C to 100^@ C ? Specific latent heat of ice = 336 J g^(-1) , specific heat capacity of water = 4.2 J g^(-1)K^(-1)

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) ]

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) .

How much boiling water at 100^@ C is needed to melt 2 kg of ice so that the mixture, which is all water, is at 0^@ C ? Given : specific heat capacity of water = 4.2 J g^(-1) K^(-1) , specific latent heat of ice = 336 J g^(-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)

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 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 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) ).

200 g of ice at 0^@ C converts into water at 0^@ C in 1 minute when heat is supplied to it at a constant rate. In how much time 200 g of water at 0^@ C will change to 20^@ C ? Take specific latent heat of ice = 336 J g^(-1) .

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)

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