$10 \, g$ of ice at $0 \, ^oC$ is mixed with $m \, g$ of water at $50 \, ^oC$. What is the minimum value of $m$ (in $g$) so that the ice melts completely? (Given: $L_f = 80 \, cal/g$ and $S_W = 1 \, cal/g \cdot ^oC$)

$1\, kg$ of ice at $-10\,^{\circ}C$ is mixed with $4.4\, kg$ of water at $30\,^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Specific heat of ice = $2100\, J/kg\cdot K$,specific heat of water = $4200\, J/kg\cdot K$,latent heat of fusion of ice = $3.36 \times 10^5\, J/kg$)

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 $= 540 \ cal/g$)

$1 \text{ kg}$ 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 $^{\circ} \text{C}$)

$A$ vessel contains $110 \, g$ of water. The heat capacity of the vessel is equivalent to $10 \, g$ of water. The initial temperature of water in the vessel is $10^{\circ}C$. If $220 \, g$ of hot water at $70^{\circ}C$ is poured into the vessel,the final temperature,neglecting radiation loss,will be........ $^{\circ}C$

$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)