Heat given to a system is $35 \ J$ and work done by the system is $15 \ J$. The change in the internal energy of the system will be ..... $J$.

When unit mass of water boils to become steam at $100\,^{\circ}C$,it absorbs $Q$ amount of heat. The densities of water and steam at $100\,^{\circ}C$ are $\rho_1$ and $\rho_2$ respectively and the atmospheric pressure is $p_0$. The increase in internal energy of the water is:

$A$ system is taken from state $A$ to state $B$ along two different paths. The heat absorbed and work done by the system along these two paths are $Q_1, Q_2$ and $W_1, W_2$ respectively,then

In a thermodynamic process,the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $30 \,J$ of heat and $10 \,J$ of work is done on the gas. If the initial internal energy of the gas was $10 \,J$,then the final internal energy will be: (in $\,J$)

Write the sign convention for internal energy in thermodynamics.

$A$ 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.