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A metal ball of specific gravity 4.5 and...

A metal ball of specific gravity `4.5` and specific heat `0.1 cal//gm-"^(@)C` is placed on a large slab of ice at `0^(@)C`. Half of the ball sinks in the ice. The initial temperature of the ball is:- (Latent heat capacity of ice `=80 cal//g`, specific gravity of ice `=0.9)`

A

`100"^(@)C`

B

`90"^(@)C`

C

`80"^(@)C`

D

`70"^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`V xx 4.5 xx 10^(3) xx 0.1 xx T = (V)/(2) xx 0.9 xx 10^(3) xx 80`
`T = (9 xx 800)/(2xx4.5 xx 11) = 80^(@)C`
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Knowledge Check

  • The specific heat of ice at 0^(@)C is

    A
    zero
    B
    infinity
    C
    in between zero to infinity
    D
    negative values

  • 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

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