$540$ 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 ...... $cal$.

An electric heater supplies heat to a system at a rate of $100 \text{ W}$. If the system performs work at a rate of $75 \text{ J/s}$, then the rate at which internal energy increases will be: (in $\text{ W}$)

The $p-V$ diagram of a system undergoing thermodynamic changes is as shown in the figure. The work done by the system in going from $A \rightarrow B \rightarrow C$ is $30 \,J$. If $68 \,J$ of heat is given to the system,then the change in the internal energy of the system between $A$ and $C$ is (in $\,J$)

The heat supplied to a gas at a constant pressure of $5 \times 10^5 \ Pa$ is $1000 \ kJ$. If the volume of gas changes from $1 \ m^3$ to $2.5 \ m^3$,then the change in internal energy of the gas is (in $kJ$)

$A$ system is given $300$ calories of heat and it does $600$ joules of work. How much does the internal energy of the system change in this process? $(1 \text{ calorie} = 4.18 \text{ joules})$

Given that $1\,g$ of water in liquid phase has volume $1\,cm^3$ and in vapour phase $1671\,cm^3$ at atmospheric pressure and the latent heat of vaporization of water is $2256\,J/g$,the change in the internal energy in joules for $1\,g$ of water at $373\,K$ when it changes from liquid phase to vapour phase at the same temperature is ....... $J$.