When unit mass of water boils to become steam | vedclass

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(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q - W$.The work done $W$ by the system during t

Vedclass · Accepted Answer

$Q + p_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)$

Vedclass · Answer

(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q - W$.The work done $W$ by the system during the phase change at constant atmospheric pressure $p_0$ is $W = p_0 \Delta V$.For a unit mass $(m = 1)$,the volume of water is $V_1 = \frac{1}{\rho_1}$ and the volume of steam is $V_2 = \frac{1}{\rho_2}$.Thus,the work done is $W = p_0 (V_2 - V_1) = p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right)$.Substituting this into the first law equation: $\Delta U = Q - p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) = Q + p_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)$.

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:

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