Steam of mass $60 \ g$ at a temperature $100^{\circ} C$ is mixed with water of mass $360 \ g$ at a temperature $40^{\circ} C$. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is $540 \ cal \ g^{-1}$ and specific heat capacity of water is $1 \ cal \ g^{-1} {}^{\circ} C^{-1}$)

$A$ massless spring $(k = 800\, N/m)$,attached with a mass $(500\, g)$ is completely immersed in $1\, kg$ of water. The spring is stretched by $2\, cm$ and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass $= 400\, J/kg\, K$,specific heat of water $= 4184\, J/kg\, K$)

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 $= 540 \ cal/g$)

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

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

Two liquids $A$ and $B$ are at temperatures of $75^{\circ} C$ and $15^{\circ} C$. Their masses are in the ratio of $2: 3$ and their specific heats in the ratio $3: 4$. If these two liquids are mixed,then the resulting temperature will be $...^{\circ} C$.