The amount of heat needed to heat $200 \ g$ of ice at $-10^{\circ}C$ to convert it into water at $30^{\circ}C$ is:
Specific heat capacity of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$
Specific heat capacity of water $= 4186 \ J \ kg^{-1} \ K^{-1}$
Latent heat of fusion of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ (in $J$)

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 $4.2\, kJ/(kg \cdot K)$]

The pressure and volume of saturated water vapour are $P$ and $V$ respectively. It is compressed isothermally such that the volume becomes $V/2$. The final pressure will be

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

$A$ waterfall is $84 \ m$ high. If half of the potential energy of the falling water gets converted to heat,the rise in temperature of the water will be: (in $^\circ C$)

How much energy will be released by cooling $1\,kg$ of water up to a temperature of $10\,^{\circ}C$?