$A$ beaker contains $200\,g$ of water. The heat capacity of the beaker is equal to that of $20\,g$ of water. The initial temperature of water in the beaker is $20\,^{\circ}C$. If $440\,g$ of hot water at $92\,^{\circ}C$ is poured in it,the final temperature (neglecting radiation loss) will be nearest to ........ $^{\circ}C$.

$4 \text{ g}$ 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 $\text{ g}$)

$A$ piece of metal of $850 \text{ K}$ is dropped into $1 \text{ kg}$ of water at $300 \text{ K}$. If the equilibrium temperature of the mixture is $350 \text{ K}$, then the heat capacity of the metal expressed in $\text{J/K}$ is (Specific heat of water = $4200 \text{ J/kg} \cdot \text{K}$)

$5\,g$ of ice at $0\,^{\circ}C$ is dropped in a beaker containing $20\,g$ of water at $40\,^{\circ}C$. The final temperature will be ........ $^{\circ}C$.

$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is

$A$ $10 \ W$ 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: