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

When $100 \ g$ of boiling water at $100^{\circ} C$ is added into a calorimeter containing $300 \ g$ of cold water at $10^{\circ} C$, the temperature of the mixture becomes $20^{\circ} C$. Then, a metallic block of mass $1 \ kg$ at $10^{\circ} C$ is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes $19^{\circ} C$. What is the specific heat of the metal in $C$.$G$.$S$. units?

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 $cal/g\,^{\circ}C.$

$A$ glass beaker contains $200 \,g$ of carbonated water initially at $20^{\circ} C$. How much ice should be added to obtain the final temperature of $0^{\circ} C$ with all ice melted, if the initial temperature of ice is $-10^{\circ} C$ (in $\,g$)? Neglect the heat capacity of the glass.
[Take, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]

The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ of ice at $0^{\circ} C$ is placed in $200 \, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of the water reaches $0^{\circ} C$ is close to (in grams):

The time taken for a calorimeter containing $75 \ g$ of water at $62^{\circ} C$ to cool to $58^{\circ} C$ is $9 \ minutes$. When the calorimeter contains $105 \ g$ of water,it takes $12 \ minutes$ to cool from $62^{\circ} C$ to $58^{\circ} C$. The water equivalent of the calorimeter is $.........$ (in $g$)