If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

Steam at $100^{\circ} C$ is passed into $1 \ kg$ of water contained in a calorimeter of water equivalent $0.2 \ kg$ at $9^{\circ} C$ until the temperature of the calorimeter and water in it increases to $90^{\circ} C$. The mass of steam condensed in $kg$ is nearly (specific heat of water $= 1 \ cal/g^{\circ} C$, latent heat of vaporisation $= 540 \ cal/g$)

$2 \; g$ 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 $cal/g$?

When $x \text{ grams}$ of steam at $100^{\circ}C$ is mixed with $y \text{ grams}$ of ice at $0^{\circ}C$,we obtain $(x + y) \text{ grams}$ of water at $100^{\circ}C$. What is the ratio $y/x$?

$A$ small quantity,mass $m$,of water at a temperature $\theta$ (in $^oC$) is poured onto a large mass $M$ of ice which is at its melting point. If $c$ is the specific heat of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is given by

$A$ calorimeter contains $0.2\, kg$ of water at $30\,^{\circ}C$. $0.1\, kg$ of water at $60\,^{\circ}C$ is added to it. The mixture is well stirred and the resulting temperature is found to be $35\,^{\circ}C$. The thermal capacity of the calorimeter is .......... $J/K$.