$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

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 $g$)

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

$1 \text{ kg}$ of steam at $150^{\circ} \text{C}$ is passed from a steam chamber into a copper coil immersed in $20 \text{ L}$ of water. The steam condenses in the coil and is returned to the steam chamber as water at $90^{\circ} \text{C}$. The latent heat of steam is $540 \text{ cal g}^{-1}$,and the specific heat of steam is $1 \text{ cal g}^{-1} {}^{\circ} \text{C}^{-1}$. The rise in temperature of the water is: (in $^{\circ} \text{C}$)

We have half a bucket $(6 \text{ litre})$ of water at $20^{\circ}C$. If we want water at $40^{\circ}C$,how much steam at $100^{\circ}C$ should be added to it?

$10\; g$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings,the temperature (in $^{\circ} C$) of water in the tumbler becomes nearly $(L_f = 80\; cal/g)$.