$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)

  • A
    $2.6$
  • B
    $2$
  • C
    $4$
  • D
    $3.2$

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Similar Questions

Steam at $100^{\circ} C$ is passed into $114 \ g$ of water at $30^{\circ} C$. The mass of water present in the mixture when the temperature of the mixture becomes $70^{\circ} C$ is (Latent heat of steam $= 540 \ cal \ g^{-1}$; specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

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$A$ $10 \ W$ electric heater is used to heat a container filled with $0.5 \ kg$ of water. It is found that the temperature of the water and the container rises by $3 \ K$ in $15 \ min$. The container is then emptied, dried, and filled with $2 \ kg$ of oil. The same heater now raises the temperature of the container-oil system by $2 \ K$ in $20 \ min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \ J \ kg^{-1} \ K^{-1}$, the specific heat of oil in the same unit is equal to:

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