$A$ tap supplies water at $10\,^{\circ}C$ and another tap at $100\,^{\circ}C$. How many $kg$ of hot water must be taken so that we get $20\, kg$ of water at $35\,^{\circ}C$?

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

Steam at $100\,^{\circ}C$ is passed into $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]

$1\, kg$ of ice at $-10\,^{\circ}C$ is mixed with $4.4\, kg$ of water at $30\,^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Specific heat of ice = $2100\, J/kg\cdot K$,specific heat of water = $4200\, J/kg\cdot K$,latent heat of fusion of ice = $3.36 \times 10^5\, J/kg$)

$200 \, g$ of a solid ball at $20 \, ^\circ C$ is dropped in an equal amount of water at $80 \, ^\circ C$. The resulting temperature is $60 \, ^\circ C$. This means that the specific heat of the solid is

$A$ kettle with $2\, L$ of water at $27\, ^oC$ is heated by a coil heater with a power of $1\, kW$. The heat is lost to the atmosphere at a constant rate of $160\, J/s$ when its lid is open. In how much time will the water be heated to $77\, ^oC$ (specific heat of water $= 4.2\, kJ/(kg \cdot K)$) with the lid open?