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How many gram of ice at -14^@C are neede...

How many gram of ice at `-14^@C` are needed to cool 200 gm of water from `25^@C` to `10^@C`. Specific heat of ice = 0.5 cal/gm `^@C` and latent heat of ice = 80 cal/gm.

Answer

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1 g of of steam at 100^@C can melt how much ice at 0^@C ? Latent heat of ice = 80 cal//g and latent heat of steam = 540 cal//g

What is the amount of work done to convert 40 g of ice at -10^(@)C into steam at 100^(@)C ? Specific heat of ice = 0.5 cal cdot g^(-1) cdot ^(@)C^(-1) , latent heat of fusion of ice =80 cal cdot g^(-1) , latent heat of steam = 540 cal cdot g^(-1), J = 4.2 xx 10^(7) erg cdot cal^(-1) .

Knowledge Check

  • 540 gm of ice at 0^@ is mixed with 540 gm of water at 80^@C . The final temperature of the mixture is

    A
    `0^@C`
    B
    `40^@C`
    C
    `80^@`
    D
    less than`0^@`C
  • 50 gm ice at 0^@C is dropped into 50 gm of water at 20^@C . The final temperature of the mixture will be

    A
    `-10^@C`
    B
    `-30^@C`
    C
    `0^@C`
    D
    `10^@C`
  • How much heat is required to raise the temperature of 100 gm of water from 30^@C to its boiling point?

    A
    27.78
    B
    35
    C
    38
    D
    25
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    How much heat is required to rise temperature from -8^@C to 50^@C of 10gm ice. Specific heat of ice 0.5 calorie gm ^-1^@C^ - 1 .

    To rise temperature of 30 gm water from 20^@C to 50^@C how much heat is required?

    5 gm of ice at -10^@C are mixed with 20 gm of water at 39^@C . Will whole of the ice melt? If so, what is the final temperature of the mixture? Specific heat capacity of ice = 0.5 cal gm^-1 "^@C^-1 and latent heat of fusion of ice = 80 cal gm^-1 , specific heat capacity of water = 1cal gm^-1 "^@C^-1 .

    1 gm of ice at 0^@C is added to 5 gm of water at 10^@C . What will be the final temperature of the mixture?

    10g of water at 30^(@)C and 5g of ice at -20^(@) C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice= 0.5 cal g^(-1) (""^(0)C)^(-1) and latent heat of fusion of ice =80 cal g^(-1)