Given m=1 kg, L v = 2257 k J / k g L_{v} = 2257\ kJ/kg Change in volume at constant pressure and temp: Δ V = V 2 − V 1 = 1.671 − 0.001 \Delta V = V_{2} - V_{1} = 1.671 - 0.001 Δ V = 1.670 m 3 \Delta V = 1.670{\text{\ }m}^{3} Using first law of thermodynamics Δ Q = Δ U + W \Delta Q = \Delta U + W Using W = P Δ V a n d Δ Q = m L v W = P\Delta V\ and\ {\Delta Q = mL}_{v} m L v = Δ U + ( 1 × 10 5 ) ( 1.670 ) {mL}_{v} = \Delta U + \left( 1 \times 10^{5} \right)(1.670) Δ U = ( 2257 − 167 ) 10 3 \Delta U = (2257 - 167)10^{3} Δ U = 2090 k J \Delta U = 2090\text{\ }kJ

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Question 1:

1 kg of water at 100 °C is converted into steam at 100 °C by boiling at atmospheric pressure. The volume of water changes from 1.00×10^-3 m³ as a liquid to 1.671 m³ as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporization = 2257 kJ/kg, Atmospheric pressure = 1 × 10⁵ Pa)

$+ 2476\text{\ }kJ$