If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively,then the first law of thermodynamics can be written as

Density of water at $4 ^\circ C$ and $20 ^\circ C$ are $1000 \ kg/m^3$ and $998 \ kg/m^3$ respectively. The increase in internal energy of $4 \ kg$ water when it is heated from $4 ^\circ C$ to $20 ^\circ C$ is . . . . . . $J$. (Specific heat capacity of water $= 4.2 \ kJ/kg \cdot K$ and $1$ atmospheric pressure $= 10^5 \ Pa$)

The $P-V$ diagram of a system undergoing a thermodynamic transformation is shown in the figure. The work done on the system in going from $A \to B \to C$ is $50 \, J$ and $20 \, cal$ of heat is given to the system. The change in internal energy between $A$ and $C$ is ...... $J$.

When $1500 \ J$ of heat is supplied to a gas at a constant pressure of $2.5 \times 10^{5} \ N/m^{2}$,the volume increases by $2.5 \times 10^{-3} \ m^{3}$. The increase in the internal energy of the gas is ..... $J$?

$A$ given quantity of gas is taken from $A$ to $C$ in two ways: $a$) directly from $A \rightarrow C$ along a straight line,and $b$) in two steps,from $A \rightarrow B$ and then from $B \rightarrow C$. The work done and heat absorbed along the direct path $A \rightarrow C$ are $200 \ J$ and $280 \ J$ respectively. If the work done along the path $A \rightarrow B \rightarrow C$ is $80 \ J$,find the heat absorbed along this path.

For a thermodynamic process $\delta Q = -50 \text{ calorie}$ and $W = -20 \text{ calorie}$. If the initial internal energy is $-30 \text{ calorie}$,then the final internal energy will be ....... $\text{calorie}$.