The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ of ice at $0^{\circ} C$ is placed in $200 \, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of the water reaches $0^{\circ} C$ is close to (in grams):

  • A
    $61.7$
  • B
    $63.8$
  • C
    $69.3$
  • D
    $64.6$

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Steam is passed into $22 \, g$ of water at $20^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90^{\circ}C$ is ........ $g$. (Latent heat of steam is $540 \, cal/g$)

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Two different rigid boxes are placed on a table,each containing a different gas. Box $A$ contains $1 \text{ mole}$ of nitrogen gas at temperature $T_0$,and box $B$ contains $1 \text{ mole}$ of helium gas at temperature $(7/3) T_0$. If these two boxes are brought into thermal contact,heat exchange occurs until a final equilibrium temperature $T_f$ is reached. The final temperature $T_f$ is:

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$100 \, g$ of water is supercooled to $-10 \, ^\circ C$. At this point,due to some disturbance,some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $[S_W = 1 \, cal \, g^{-1} \, ^\circ C^{-1}$ and $L_{fusion} = 80 \, cal \, g^{-1}]$

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