$A$ 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

Steam at $100^{\circ}C$ is added slowly to $1400 \, \text{g}$ of water at $16^{\circ}C$ until the temperature of the water is raised to $80^{\circ}C$. The mass of steam required to do this is $(L_V = 540 \, \text{cal/g})$ ........... $\text{g}$

The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$. The water is initially at a different temperature $T_W$. Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium,with no loss of heat to the surroundings. The final temperature of the mixture will be

$A$ 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, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]

If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

$A$ metal ball of mass $0.1 \, kg$ is heated up to $500 \, ^\circ C$ and dropped into a vessel of heat capacity $800 \, J K^{-1}$ containing $0.5 \, kg$ of water. The initial temperature of the water and the vessel is $30 \, ^\circ C$. The approximate percentage increment in the temperature of the water is ........ $\%$. [Specific heat capacities of water and metal are $4200 \, J kg^{-1} K^{-1}$ and $400 \, J kg^{-1} K^{-1}$ respectively.]