When $x \text{ grams}$ of steam at $100^{\circ}C$ is mixed with $y \text{ grams}$ of ice at $0^{\circ}C$,we obtain $(x + y) \text{ grams}$ of water at $100^{\circ}C$. What is the ratio $y/x$?

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 $...^{\circ} C$.

In an experiment on the specific heat of a metal,a $0.20 \; kg$ block of the metal at $150 \; ^{\circ}C$ is dropped in a copper calorimeter (of water equivalent $0.025 \; kg$) containing $150 \; cm^{3}$ of water at $27 \; ^{\circ}C$. The final temperature is $40 \; ^{\circ}C$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible,is your answer greater or smaller than the actual value for the specific heat of the metal?

$5 \ kg$ of water at $20^{\circ} C$ is added to $10 \ kg$ of water at $60^{\circ} C$. Neglecting the heat capacity of the vessel and other losses,the resultant temperature will be nearly: (in $^{\circ} C$)

$A$ $2100 W$ continuous flow geyser (instant geyser) has water inlet temperature $= 10^{\circ}C$ while the water flows out at the rate of $20\,g/s$. The outlet temperature of water must be about ....... $^{\circ}C$.

$A$ $10 \ W$ electric heater is used to heat a container filled with $0.5 \ kg$ of water. It is found that the temperature of the water and the container rises by $3 \ K$ in $15 \ min$. The container is then emptied,dried,and filled with $2 \ kg$ of oil. The same heater now raises the temperature of the container-oil system by $2 \ K$ in $20 \ min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \ J \ kg^{-1} \ K^{-1}$,the specific heat of oil in the same unit is equal to: