When a system is taken from state $i$ to state $f$ along the path $iaf$,it is found that $Q = 50 \ cal$ and $W = 20 \ cal$. Along the path $ibf$,$Q = 36 \ cal$. The work done $W$ along the path $ibf$ is ....... $cal$.

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : If $dQ$ and $dW$ represent the heat supplied to the system and the work done on the system respectively,then according to the first law of thermodynamics $dQ = dU - dW$.
Reason $R$ : The first law of thermodynamics is based on the law of conservation of energy.
In the light of the above statements,choose the correct answer from the options given below:

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 ....... $\text{calorie}$.

The first law of thermodynamics states that:

$A$ 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

Calculate the change in internal energy when $1 \,g$ of water is converted into steam at atmospheric pressure $(1.013 \times 10^{5} \,Pa)$. Given the latent heat of vaporisation is $2256 \,J/g$, the volume of $1 \,g$ of water is $1 \,cm^{3}$, and the volume of $1 \,g$ of steam is $1671 \,cm^{3}$.