Question

The pressure of water is increased from 100 kPa to 900 kPa by a pump. The temperature of water also increases by 0.15°C. The density of water is 1 kg/L and its specific heat is cp = 4.18 kJ/kg⋅°C. The enthalpy change of the water during this process is _____ kJ/kg. Solve this problem using appropriate software.

Answer 1

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

$\Delta H = \frac{\Delta P}{\rho} + C_{p} \Delta T$

Where

$\Delta H$ is the enthalpy change

$\Delta P$ is the change in pressure

$\rho$ is the density

$C_{p}$ is the specific heat

and $\Delta T$ is the temperature change

From the question

$\Delta P$ = 900 kPa - 100 k Pa

$\Delta P$ = 800 kPa

$\rho$ = 1 kg/L = 1kg/dm³ = 1000 kg/m³

$C_{p}$ = 4.18 kJ/kg.°C

$\Delta T$ = 0.15 °C

∴$\Delta H = \frac{\Delta P}{\rho} + C_{p} \Delta T$

$\Delta H = \frac{800}{1000} +4.18 \times 0.15$

$\Delta H = 0.8 + 0.627$

$\Delta H = 1.4 kJ /kg$

Hence, the enthalpy change of the water during the process is 1.4 kJ/kg