$A$ block of ice at $-20\,^oC$ having a mass of $2\,kg$ is added to $3\,kg$ of water at $15\,^oC$. Neglecting heat losses and the heat capacity of the container,what is the final state of the system?

Three bodies of the same material and having masses $m, m$ and $3m$ are at temperatures $40^{\circ} C, 50^{\circ} C$ and $60^{\circ} C$ respectively. If the bodies are brought in thermal contact, the final temperature will be (in $^{\circ} C$)

Steam at $100^{\circ} C$ is passed into $114 \ g$ of water at $30^{\circ} C$. The mass of water present in the mixture when the temperature of the mixture becomes $70^{\circ} C$ is (Latent heat of steam $= 540 \ cal \ g^{-1}$; specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

$A$ liquid of specific heat $0.8 \ cal / g^{\circ} C$ at temperature $60^{\circ} C$ is mixed with another liquid of the same mass having temperature $45^{\circ} C$. If the temperature of the mixture is $53^{\circ} C$,then the specific heat (in $cal / g^{\circ} C$) of the second liquid is:

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 $= 540 \ cal/g$)

$A$ $10 \ W$ 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: