Question

When unit mass of water boils to become steam at ${100}^{\circ }\text{C}$, it absorbs $Q$ amount of heat. The densities of water and steam at ${100}^{\circ }\text{C}$ are ${\rho }_1$and ${\rho }_2$ respectively, and the atmospheric pressure is $p_0$. The increase in the internal energy of the water is
(correct answer + 1, wrong answer - 0.25)

• A. $Q$
• B. $Q+p_0(\frac{1}{{\rho }_1}-\frac{1}{{\rho }_2})$
• C. $Q+p_0(\frac{1}{{\rho }_2}-\frac{1}{{\rho }_1})$
• D. $Q-p_0(\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2})$

Answer 1

The correct option is B $Q+p_0(\frac{1}{{\rho }_1}-\frac{1}{{\rho }_2})$

The work done by unit mass of water during expansion to form steam at constant pressure $p_0$ is
$W=p_0\Delta V=p_0(V_{\text{steam}}-V_{\text{water}})$
$=p_0(\frac{1}{{\rho }_2}-\frac{1}{{\rho }_1})$
From first law of thermodynamics:-
$\Delta U=Q-W=Q-p_0(\frac{1}{{\rho }_2}-\frac{1}{{\rho }_1})$