Answer:
The enthalpy change of the water during the process is 1.4 kJ/kg
Explanation:
The enthalpy change of the water during the process can be determined from
[tex]\Delta H = \frac{\Delta P}{\rho} + C_{p} \Delta T[/tex]
Where
[tex]\Delta H[/tex] is the enthalpy change
[tex]\Delta P[/tex] is the change in pressure
[tex]\rho[/tex] is the density
[tex]C_{p}[/tex] is the specific heat
and [tex]\Delta T[/tex] is the temperature change
From the question
[tex]\Delta P[/tex] = 900 kPa - 100 k Pa
[tex]\Delta P[/tex] = 800 kPa
[tex]\rho[/tex] = 1 kg/L = 1kg/dm³ = 1000 kg/m³
[tex]C_{p}[/tex] = 4.18 kJ/kg.°C
[tex]\Delta T[/tex] = 0.15 °C
∴[tex]\Delta H = \frac{\Delta P}{\rho} + C_{p} \Delta T[/tex]
[tex]\Delta H = \frac{800}{1000} +4.18 \times 0.15[/tex]
[tex]\Delta H = 0.8 + 0.627\\[/tex]
[tex]\Delta H = 1.4 kJ /kg[/tex]
Hence, the enthalpy change of the water during the process is 1.4 kJ/kg
