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

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

Step-by-step explanation:

The enthalpy change of the water during the process can be determined from

`\Delta H = (\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 = (\Delta P)/(\rho) + C_(p) \Delta T`

`\Delta H = (800)/(1000) +4.18 * 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