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 $y/x$?

$A$ calorimeter has a heat capacity of $100 \ J/K$ and is at $30^{\circ}C$. $100 \ g$ of water at $40^{\circ}C$ (specific heat capacity $4200 \ J/kg \cdot K$) is poured into the calorimeter. What is the final temperature of the mixture in the calorimeter in $^{\circ}C$?

$A$ metal ball of mass $100 \ g$ at $20^{\circ} C$ is dropped in $200 \ ml$ of water at $80^{\circ} C$. If the resultant temperature is $70^{\circ} C$,then the ratio of specific heat of the metal to that of water is

$10 \, g$ of ice at $-20^{\circ} C$ is kept in a calorimeter containing $10 \, g$ of water at $10^{\circ} C$. The specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain ..........

$100 \text{ g}$ 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 $^{\circ}C$)

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be: