$A$ thermally insulated vessel contains $150\, g$ of water at $0\, ^oC$. Then the air from the vessel is pumped out adiabatically. $A$ fraction of water turns into ice and the rest evaporates at $0\, ^oC$ itself. The mass of evaporated water will be closest to ....... $g$ (Latent heat of vaporization of water $= 2.10 \times 10^6\, J/kg$ and Latent heat of fusion of water $= 3.36 \times 10^5\, J/kg$).

$10 \, g$ of ice at $0^{\circ} C$ is kept in a calorimeter of water equivalent $10 \, g$. How much heat (in $cal$) should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat)

When the pressure on water is increased,the boiling temperature of water as compared to $100^{\circ}C$ will be:

Consider a rectangular sheet of solid material of length $l=9 \ cm$ and width $d=4 \ cm$. The coefficient of linear expansion is $\alpha=3.1 \times 10^{-5} \ K^{-1}$ at room temperature and one atmospheric pressure. The mass of the sheet is $m=0.1 \ kg$ and the specific heat capacity is $C_v=900 \ J \ kg^{-1} K^{-1}$. If the amount of heat supplied to the material is $8.1 \times 10^2 \ J$,then the change in area of the rectangular sheet is:

An electric heater with a constant heat supply rate is used to convert a certain amount of liquid ammonia to saturated vapour at high pressure. The heater takes $14 \text{ minutes}$ to bring the liquid at $15^{\circ}C$ to the boiling point of $50^{\circ}C$ and $92 \text{ minutes}$ to convert the liquid at the boiling point wholly to vapour. If the specific heat capacity of liquid ammonia is $4.9 \text{ kJ/kg K}$, the latent heat of vaporisation of ammonia in $\text{kJ/kg}$ is:

The temperature of a body is given by $T = 6t^2 + 4$. The rate of change of temperature with respect to time at $t = 10 \ s$ is ........ $K/s$. ($T$ is in $Kelvin$ and $t$ is in $seconds$)