$4 \text{ g}$ of steam at $100^{\circ} C$ is added to $20 \text{ g}$ of water at $46^{\circ} C$ in a container of negligible mass. Assuming no heat is lost to the surroundings, the mass of water in the container at thermal equilibrium is. (Latent heat of vaporisation $= 540 \text{ cal/g}$, Specific heat of water $= 1 \text{ cal/g}^{\circ} C$):- (in $\text{ g}$)

  • A
    $18$
  • B
    $20$
  • C
    $22$
  • D
    $24$

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Similar Questions

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

$A$ $10 \ W$ 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:

$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

$A$ $2100 W$ continuous flow geyser (instant geyser) has water inlet temperature $= 10^{\circ}C$ while the water flows out at the rate of $20\,g/s$. The outlet temperature of water must be about ....... $^{\circ}C$.

$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

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