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A calorimeter of water equivalent 20 g...

A calorimeter of water equivalent 20 g contains 180 g of water at `25^(@)C` . 'm' grams of steam at `100^(@)C` is mixed in it till the temperature of the mixture is `31^(@)C` . The value of m is close to (Laten heat of water = 540 `cal g^(-1)` , specific heat of water = 1 cal `g^(-1) ""^(@)C^(-1))`

A

2

B

2.6

C

4

D

3.2

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Specific heat of water is 1 cal g^(-1) .^(@)C^(-1)

Specific heat of water is 1"cal"//g//^(@)C .

Knowledge Check

  • 540 g of ice at 0^(@)C is mixed with 540 g of water at 80^(@)C . The final temperature of the mixture is

    A
    `0^(@)C`
    B
    `40^(@)C`
    C
    `80^(@)C`
    D
    Less than `0^(@)C`

  • 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
    None of these

  • A calorimeter contains 70.2 g of water at 15.3^@C . IF 143.7 g of water at 36.5^@C is mixed with it, the common temperature becomes 28.7^@C . The water equivalent of a calorimeter is

    A
    15.6 g
    B
    9.4 g
    C
    6.3 g
    D
    13.4 g

    • Similar Questions

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    20 g of steam at 100^@ C is passes into 100 g of ice at 0^@C . Find the resultant temperature if latent heat if steam is 540 cal//g ,latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g^@C .

    50 g of steam at 100^@ C is passed into 250 g of at 0^@ C . Find the resultant temperature (if latent heat of steam is 540 cal//g , latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g -.^@ C ).

    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 coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t_(0) seconds to make coffee at 70^(@)C when the initial temperature of the mixture is 25^(@)C . The value of t_(0) is close to (ratio of latent heat of evaporation to specific heat of water is 540^(@)C ) and specific heat of the mixture can taken to be the same as that of water)

    The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15^@ C. Some amount of ice, initially at -10^@ C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal g^(-1) ""^(@) C^(-1), specific heat of water = 1.0 cal g^(-1) ""^(@) C^(-1) and latent heat of melting of ice = 80 cal g'^(-1) )?

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