$540$ calories of heat convert $1$ cubic centimeter of water at $100^{\circ}C$ into $1671$ cubic centimeter of steam at $100^{\circ}C$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly ...... $cal$.

Consider $1 \,kg$ of liquid water undergoing a change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$,the vapour pressure is $1.01 \times 10^5 \,N m^{-2}$ and the latent heat of vaporization is $22.6 \times 10^5 \,J kg^{-1}$. The density of liquid water is $10^3 \,kg m^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m^{-3}$. The change in internal energy in this phase change is nearly ............ $J kg^{-1}$.

The first law of thermodynamics states that:

If the heat given to a system is $35 \ J$ and the work done on the system is $15 \ J$,then the change in the internal energy of the system is ... $J$?

$1\;g$ of water,of volume $1\;cm^{3}$ at $100^{\circ}C$,is converted into steam at the same temperature under normal atmospheric pressure $(=1 \times 10^{5}\;Pa)$. The volume of steam formed equals $1671\;cm^{3}$. If the specific latent heat of vaporisation of water is $2256\;J/g$,the change in internal energy is.....$J$.

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then the relation between $\Delta U_1$ and $\Delta U_2$ is