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One kilogram of ice at 0^(@)C is mixed w...

One kilogram of ice at `0^(@)C` is mixed with one kilogram of water at `80^(@)C`. The final temperature of the mixture is
(Take : specific heat of water `=4200 J kg^(-1) K^(-1)`, latent heat of ice `= 336 kJ//kg^(-1)`)

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Knowledge Check

  • 336 g of ice at 0^@C is mixed with 336 g of water at 80^@C . What is the final temperature of the mixture?

    A
    `0^@C`
    B
    `40^@C`
    C
    `80^@C`
    D
    `85^@C`

  • 1kg of ice at 0^(@)C is mixed with 1kg of water at 10^(@)C . The resulting temperature will be

    A
    between `0^(@)C and 10^(@)C`
    B
    greater than `10^(@)C`
    C
    less than `0^(@)C`
    D
    equal to `0^(@)C`

  • One kg of ice at 0^(@)C is mixed with 1 kg of water at 10^(@)C . The resulting temperature will be

    A
    between `0^(@)C` and `10^(@)C`
    B
    `0^(@)C`
    C
    less than `0^(@)C`
    D
    greater than `0^(@)C`

    • Similar Questions

    Explore conceptually related problems

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    10 g of ice of 0^(@)C is mixed with 100 g of water at 50^(@)C in a calorimeter. The final temperature of the mixture is [Specific heat of water = 1 cal g^(-1).^(@)C^(-1) , letent of fusion of ice = 80 cal g^(-1) ]

    If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

    50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

    1g of ice at 0^(@)C is added to 5g of water at 10^(@)C . If the latent heat is 80cal/g, the final temperature of the mixture is

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