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1g of ice at 0^@ C is mixed 1 g of steam...

`1g` of ice at `0^@ C` is mixed `1 g` of steam at `100^@ C`. The mass of water formed is

A

`1.33 g`

B

`1 g`

C

`0.133 g`

D

`13.3 g`

Text Solution

Verified by Experts

The correct Answer is:
A
Here the resultant temperature is `100^@ C`
`m^(|)` is mass of the steam condensed
`m^(|)L_(v) =m_(ice)L_(f)+m_(ice)S_(water) xx Delta theta`
:. Warer formed `= 1g + m^(|)`.
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Knowledge Check

  • If 10 g of ice at 0^(@)C is mixed with 10 g of water at 40^(@)C . The final mass of water in mixture is (Latent heat of fusion of ice = 80 cel/g, specific heat of water =1 cal/g""^(@)C )

    A
    10 g
    B
    15 g
    C
    18 g
    D
    20 g

  • 1 g of ice at 0^@C is mixed with 1 g of steam at 100^@C . After thermal equilibrium is achieved, the temperature of the mixture is

    A
    `100^@C`
    B
    `55^@C`
    C
    `75^@C`
    D
    `0^@C`

  • 1 kg of ice at 0^@C is mixed with 1 kg of steam at 100^@C (i) The equilibrium temperature is 100^@C (ii) The equilibrium temperature is 0^@C (iii) The contents of mixture are : 665 g steam, 1335 g water (iv) The contents of mixture are 800 g steam, 1200 g water

    A
    (i),(iii)
    B
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    C
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    D
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