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30 kg ice at 0^@C and 20 g of steam at 1...

`30 kg` ice at `0^@C` and `20 g` of steam at `100^@C` are mixed. The composition of the resultant mixture is

A

`40 g` of water and `10 g` steam at `100^@C`

B

`10g` of ice and `40 g` of water at `0^@C`

C

`50g` of water at `100^@C`

D

`35 g` of water and `15 g` of steam at `100^@ C`

Text Solution

Verified by Experts

The correct Answer is:
A
`m_(s)L_(s) =(m_(ice)L_(ice) + m_(ice) s Delta theta)`
Where `m_(s)` = mass f steam condensed to rise temperature of ice to `100^@C` water.
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Knowledge Check

  • 5 g of ice at 0^@C and 20 g of water at 45^@C are mixed. The temperature of the mixture will be

    A
    `10^@C`
    B
    `20^@C`
    C
    `30^@C`
    D
    `40^@C`

  • 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`

  • 1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )

    A
    `50^@C`
    B
    `100^@C`
    C
    `67^@C`
    D
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