1 text{ cm}^3 of water at its boiling point a | vedclass

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(C) According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. Here, $\Delta Q$ is the total heat absorbed, $\Delta U$ is the cha

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$500$

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(C) According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. Here, $\Delta Q$ is the total heat absorbed, $\Delta U$ is the change in internal energy (energy spent in overcoming intermolecular forces), and $\Delta W$ is the work done against atmospheric pressure. Given: $\Delta Q = 540 	ext{ cal}$, $P = 1.013 	imes 10^5 	ext{ N/m}^2$, $V_1 = 1 	ext{ cm}^3 = 10^{-6} 	ext{ m}^3$, $V_2 = 1671 	ext{ cm}^3 = 1671 	imes 10^{-6} 	ext{ m}^3$, and $J = 4.19 	ext{ J/cal}$. Work done $\Delta W = P(V_2 - V_1) = 1.013 	imes 10^5 	imes (1671 - 1) 	imes 10^{-6} 	ext{ J} = 1.013 	imes 10^5 	imes 1670 	imes 10^{-6} 	ext{ J} \approx 169.17 	ext{ J}$. Converting work to calories: $\Delta W_{	ext{cal}} = \frac{169.17}{4.19} \approx 40.37 	ext{ cal}$. Internal energy change $\Delta U = \Delta Q - \Delta W_{	ext{cal}} = 540 - 40.37 \approx 500 	ext{ cal}$.

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