$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)

$0.1 \, m^3$ of water at $80 \, ^oC$ is mixed with $0.3 \, m^3$ of water at $60 \, ^oC$. The final temperature of the mixture is ........ $^oC$.

The temperature of $100 \,g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required in this process is (latent heat of steam is $540 \,cal \,g^{-1}$). (in $\,g$)

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$,specific heat of water $= 4200 \ J/kg.^{\circ}C$,latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)

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:

$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established,the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)