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Specific heat capacity of substance A is...

Specific heat capacity of substance A is `3.8Jg^(-1)K^(-1)` whereas the specific heat capacity of substance B is `0.4Jg^(-1)K^(-1)`.
How is one led to the above conclusion ?

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To solve the problem regarding the specific heat capacities of substances A and B, we can follow these steps: ### Step-by-Step Solution 1. **Understanding Specific Heat Capacity**: - Specific heat capacity (C) is defined as the amount of heat (Q) required to raise the temperature of a unit mass (m) of a substance by one degree Kelvin (or Celsius). - The formula relating these quantities is given by: \[ ...
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Specific heat capacity of substance A is 3.8Jg^(-1)K^(-1) whereas the specific heat capacity of substance B is 0.4Jg^(-1)K^(-1) . If substances A and B are liquids then which one would be more useful in car radiators ?

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Calculate the total amount of heat energy required to convert 100 g of ice at -10^@ C completely into water at 100^@ C. Specific heat capacity of ice = 2.1 J g^(-1) K^(-1) , specific heat capacity of water = 4.2 J g^(-1) K^(-1) , 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)

Calculate the heat required to convert 3 kg of ice at -12^(@)C kept in a calorimeter to steam at 100^(@)C at atmospheric pressure. Given, specific heat capacity of ice = 2100 J kg^(-1) K^(-1) specific heat capicity of water = 4186 J kg^(-1)K^(-1) Latent heat of fusion of ice = 3.35 xx 10^(5) J kg^(-1) and latent heat of steam = 2.256 xx 10^(6) J kg^(-1) .

Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32^@ C such that the final temperature is 5^@ C. Specific heat capacity of calorimeter = 0.4 J g^(-1)^@C^(-1) , specific heat capacity of water = 4.2" J "g^(-1)""^@C^(-1) , latent heat capacity of ice = 330 J g^(-1) .

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A refrigerator converts 100 g of water at 20^@ C to ice at - 10^@ C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g^(-1) K^(-1) , specific latent heat of ice is 336 J g^(-1) and the specific heat capacity of ice is 2.1 J g^(-1) K^(-1)

In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34^@ C to its freezing temperature. The specific heat capacity of water is 4.2 J g^(-1) K^(-1) . Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

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