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2 kg of ice at -15^@C is mixed with 2.5 ...

2 kg of ice at `-15^@C` is mixed with 2.5 kg of water at `26^@C` in an insulating container. If the specific heat capacities of ice and water are `0.5 cal//g^@C` and `1 cal//g^@C`, find the amount of water present in the container? (in kg nearest integer)

Text Solution

Verified by Experts

Energy relesased by water from `25^@C` to `0^@C`
`=2500xx1xx25=62500kcal`
Energy to bring ice to `0^@C`
`=2000xx(1)/(2)xx15=15000cal`
Energy used to melt ice of m gram `=m80cal`
Ice melt `m=((62500-15000)/(80))=593.75g`
So, mass of water `=(2500+593.75)g`
`=3093.75gcong3kg`
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Knowledge Check

  • 1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

    A
    the temperature of the mixture is `0_(@)C`
    B
    mixture contains `156.25` g of ice
    C
    mixture contains `843.75` g of ice
    D
    the temperature of the mixture is `15^(@)C`.

  • 200 g of ice at –20°C is mixed with 500 g of water 20°C in an insulating vessel. Final mass of water in vessel is (specific heat of ice – 0.5 cal g^(–1°)C^(–1) )

    A
    700 g
    B
    600 g
    C
    400 g
    D
    200 g

  • 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)

    A
    `0^@C`
    B
    `40^@C`
    C
    `60^@C`
    D
    less than `0^@C`