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100 g of water is supercooled to - 10^@C...

100 g of water is supercooled to - `10^@C.` At this point, due to same disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperautre of the resultant mixture and how much mass would freeze ?`[s_w = 1 cal//g// .^@ C and L_(Fusion)^(w) = 80 cal//g]`

A

`10^@ C`

B

`0^@ C`

C

`-5^@ C`

D

`-2^@ C`

Text Solution

Verified by Experts

The correct Answer is:
B
Heat loss by hot body = heat gain by cold body.
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Knowledge Check

  • 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`

  • 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

    A
    `5^(@)C`
    B
    `0^(@)C`
    C
    `-5^(@)C`
    D
    None of these

  • What will be the final temperature when 150 g of ice at 0^@C is mixed with 300 g of water at 50^@C . Specific heat of water =1 cal//g//^@C . Latent heat of fusion of ice =80 cal//g .

    A
    `7.6^@C`
    B
    `6.7^@C`
    C
    `5.8^@C`
    D
    `8.5^@C`

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