$A$ liquid of mass $M$ and specific heat $S$ is at a temperature $2t$. If another liquid of thermal capacity $1.5$ times,at a temperature of $\frac{t}{3}$ is added to it,the resultant temperature will be

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):

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

$2\,kg$ 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:

$300 \text{ g}$ 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 $^{\circ}C$)

$4 \text{ g}$ of steam at $100^{\circ} C$ is added to $20 \text{ g}$ of water at $46^{\circ} C$ in a container of negligible mass. Assuming no heat is lost to the surroundings, the mass of water in the container at thermal equilibrium is. (Latent heat of vaporisation $= 540 \text{ cal/g}$, Specific heat of water $= 1 \text{ cal/g}^{\circ} C$):- (in $\text{ g}$)