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The specific heat of water 4200 "J kg"^(...

The specific heat of water `4200 "J kg"^(-1)K^(-1)` and latent heat of ice `=3.4xx10^5 "j kg"^(-1)` . 100 grams of ice at `0^@C` is placed in 200 g of water at `25^@C`. The amount of ice that will melt as the temperature of water reaches `0^@C` is close to (in grams)

A

63.8

B

69.3

C

64.6

D

61.7

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100 g of ice at 0^@C is mixed with 0.25 kg of water at 0^@C . The net transfer of heat is______

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Knowledge Check

  • One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

    A
    `5^(@)C`
    B
    `0^(@)C`
    C
    `-5^(@)C`
    D
    None of the above

  • m_(1) gram of ice at -10^(@)C and m_(2) gram of water at 50^(@)C are mixed in insulated container. If in equilibrium state we get only water at 0^(@)C then latent heat of ice is :

    A
    `(50m_(2))/(m_(1))+5`
    B
    `(50m_(2))/(m_(1))-5`
    C
    `(50m_(2))/(m_(1))+50`
    D
    `(50m_(2))/(m_(1))-50`

  • 120 g of ice at 0^(@)C is mixed with 100 g of water at 80^(@)C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/ g-.^(@)C . The final temperature of the mixture is

    A
    `0^(@)C`
    B
    `40^(@)C`
    C
    `20^(@)C`
    D
    `10^(@)C`

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    A piece of ice (heat capacity =2100 J kg^(-1) .^(@)C^(-1) and latent heat =3.36xx10^(5) J kg^(-1) ) of mass m grams is at -5 .^(@)C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice . Water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m in gram is

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    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) ).

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