$19 \ g$ of water at $30^{\circ} C$ and $5 \ g$ of ice at $-20^{\circ} C$ are mixed together in a calorimeter. What is the final temperature of the mixture (in $^{\circ} C$)? Given specific heat of ice $= 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$.

An iron ball of mass $0.2\,kg$ is heated to $100\,^{\circ}C$ and put into a block of ice at $0\,^{\circ}C.$ If $25\,g$ of ice melts,and the latent heat of fusion of ice is $80\,cal/g,$ find the specific heat of iron in $cal/g\,^{\circ}C.$

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

Steam at $100^{\circ}C$ is passed into $1.1 \, kg$ of water contained in a calorimeter of water equivalent $0.02 \, kg$ at $15^{\circ}C$ until the temperature of the calorimeter and its contents rises to $80^{\circ}C$. The mass of the steam condensed in $kg$ is:

$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

$300 \text{ g}$ of water at $25^{\circ}C$ is added to $100 \text{ g}$ of ice at $0^{\circ}C$. The final temperature of the mixture is (in $^{\circ}C$)