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Home » of water, of volume at , is converted into steam at same temperature under normal atmospheric pressure . The volume of steam formed equals . If the specific latent heat of vaporisation of water is , the change in internal energy is, (1) (2) (3) (4)
1 \mathrm{~g} of water, of volume 1 \mathrm{~cm}^{3} at 100^{\circ} \mathrm{C}, is converted into steam at same temperature under normal atmospheric pressure \left(\therefore 1 \times 10^{5} \mathrm{~Pa}\right). The volume of steam formed equals 1671 \mathrm{~cm}^{3}. If the specific latent heat of vaporisation of water is 2256 \mathrm{~J} / \mathrm{g}, the change in internal energy is, (1) 2256 \mathrm{~J} (2) 2423 \mathrm{~J} (3) 2089 \mathrm{~J} (4) 167 \mathrm{~J}
Physics | Physics
1 \mathrm{~g} of water, of volume 1 \mathrm{~cm}^{3} at 100^{\circ} \mathrm{C}, is converted into steam at same temperature under normal atmospheric pressure \left(\therefore 1 \times 10^{5} \mathrm{~Pa}\right). The volume of steam formed equals 1671 \mathrm{~cm}^{3}. If the specific latent heat of vaporisation of water is 2256 \mathrm{~J} / \mathrm{g}, the change in internal energy is, (1) 2256 \mathrm{~J} (2) 2423 \mathrm{~J} (3) 2089 \mathrm{~J} (4) 167 \mathrm{~J}

Answer (3)
Sol.
\begin{aligned} \Delta Q &=2256 \times 1=2256 \mathrm{~J} \\ \Delta \mathrm{W} &=P\left[V_{\text {steam }}-V_{\text {water }}\right] \\ &=1 \times 10^{5}[1671-1] \times 10^{-6} \\ &=1670 \times 10^{5} \times 10^{-6} \\ &=167 \mathrm{~J} \end{aligned}
By first law of thermodynamics:
\begin{array}{l} \text { As } \Delta Q=\Delta U+\Delta W \\ 2256=\Delta U+167 \\ \Delta U=2089 \mathrm{~J} \end{array}

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