Ice at $-20\,^{\circ}C$ is added to $50\,g$ of water at $40\,^{\circ}C.$ When the temperature of the mixture reaches $0\,^{\circ}C,$ it is found that $20\,g$ of ice is still unmelted. The amount of ice added to the water was close to ........$g$ (Specific heat of ice $= 2.1\,J/g/^{\circ}C,$ Specific heat of water $= 4.2\,J/g/^{\circ}C,$ Heat of fusion of water at $0\,^{\circ}C = 334\,J/g).$

Steam at $100\,^{\circ}C$ is passed into $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]

The time taken for a calorimeter containing $75 \ g$ of water at $62^{\circ} C$ to cool to $58^{\circ} C$ is $9 \ minutes$. When the calorimeter contains $105 \ g$ of water,it takes $12 \ minutes$ to cool from $62^{\circ} C$ to $58^{\circ} C$. The water equivalent of the calorimeter is $.........$ (in $g$)

$A$ liquid of specific heat $0.8 \ cal / g^{\circ} C$ at temperature $60^{\circ} C$ is mixed with another liquid of the same mass having temperature $45^{\circ} C$. If the temperature of the mixture is $53^{\circ} C$,then the specific heat (in $cal / g^{\circ} C$) of the second liquid is:

$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

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 $g$)?