1 g of water of volume $1 ext{ cm}^3$ at $100^{\circ} ext{C}$ is converted into steam at the same temperatu

Question

1 g of water of volume $1 	ext{ cm}^3$ at $100^{\circ}	ext{C}$ is converted into steam at the same temperature under normal atmospheric pressure $\approx 1 	imes 10^5 	ext{ Pa}$. The volume of steam formed equals $1671 	ext{ cm}^3$. If the specific latent heat of vaporization of water is $2256 	ext{ J/g}$, the change in internal energy is:

Accepted Answer

According to the First Law of Thermodynamics, the change in internal energy ($\Delta U$) is given by:
$$\Delta Q = \Delta U + \Delta W \implies \Delta U = \Delta Q - \Delta W$$

1.  **Calculate Heat Absorbed ($\Delta Q$):**
    The heat absorbed during the phase change is determined by the specific latent heat of vaporization ($L$) and the mass ($m$).
    $$\Delta Q = m 	imes L$$
    $$\Delta Q = 1 	ext{ g} 	imes 2256 	ext{ J/g} = 2256 	ext{ J}$$

2.  **Calculate Work Done ($\Delta W$):**
    Work done during expansion at constant pressure is given by $\Delta W = P\Delta V$.
    *   Pressure ($P$) = $1 	imes 10^5 	ext{ Pa}$
    *   Change in volume ($\Delta V$) = $V_{	ext{steam}} - V_{	ext{water}} = 1671 	ext{ cm}^3 - 1 	ext{ cm}^3 = 1670 	ext{ cm}^3$
    *   Convert $\Delta V$ to SI units ($\text{m}^3$): $1670 	imes 10^{-6} 	ext{ m}^3$

$$\Delta W = (1 	imes 10^5 	ext{ Pa}) 	imes (1670 	imes 10^{-6} 	ext{ m}^3) = 167 	ext{ J}$$

3.  **Calculate Internal Energy Change ($\Delta U$):**
    $$\Delta U = 2256 	ext{ J} - 167 	ext{ J} = 2089 	ext{ J}$$

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