Water of volume $2 \, L$ in a container is heated with a coil of $1 \, kW$ at $27 \, ^\circ C$. The lid of the container is open and energy dissipates at a rate of $160 \, J/s$. In how much time will the temperature rise from $27 \, ^\circ C$ to $77 \, ^\circ C$? [Given: specific heat of water is $4.2 \, kJ/(kg \cdot K)$]

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

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

$A$ 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?

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$, specific heat of water $= 4200 \ J/kg.^{\circ}C$, latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice, initially at $-10^{\circ} C$, is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$, specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?