liquid water at 100C and 1 bar has internal energy (on an arbitrary scale) of 419 kj /kg and a specific volume of 1.044cm3 /g

a- what is its enthalpy?
b-the water brought to the vapor state at 200C and 800 kpa, where its enthalpy is 2838.6 kj/kg and it is specific volume is 260.79 cm3/g. calculate delta U and delta H for the process?

To answer these questions, we need to understand some basic thermodynamic concepts and equations. Specifically, the enthalpy and internal energy equations are relevant to solving these problems.

a) Enthalpy (H) is defined as the internal energy (U) plus the product of the pressure (P) and specific volume (v). Mathematically, the equation is:

H = U + Pv

In this case, the given values are:
Internal energy (U) = 419 kJ/kg
Specific volume (v) = 1.044 cm³/g
Pressure (P) = 1 bar

First, convert the pressure and specific volume to SI units:
Specific volume (v) = 1.044 cm³/g = 1.044 × 10⁻⁶ m³/kg
Pressure (P) = 1 bar = 100 kPa

Now, we can calculate the enthalpy:
H = 419 kJ/kg + (100 kPa × 1.044 × 10⁻⁶ m³/kg)
H = 419 kJ/kg + 0.1044 kJ/kg
H = 419.1044 kJ/kg

So, the enthalpy of liquid water at 100°C and 1 bar is approximately 419.1044 kJ/kg.

b) The change in internal energy (ΔU) and enthalpy (ΔH) during the process of water vaporization can be calculated using the following equations:

ΔU = U₂ - U₁
ΔH = H₂ - H₁

Given values:
Initial temperature (T₁) = 100°C
Initial pressure (P₁) = 1 bar
Final temperature (T₂) = 200°C
Final pressure (P₂) = 800 kPa
Initial internal energy (U₁) = 419 kJ/kg
Final enthalpy (H₂) = 2838.6 kJ/kg

First, convert the temperatures to Kelvin:
Initial temperature (T₁) = 100°C + 273.15 = 373.15 K
Final temperature (T₂) = 200°C + 273.15 = 473.15 K

Next, convert the pressures and specific volumes to SI units:
Initial pressure (P₁) = 1 bar = 100 kPa
Final pressure (P₂) = 800 kPa
Initial specific volume (v₁) = 1.044 cm³/g = 1.044 × 10⁻⁶ m³/kg
Final specific volume (v₂) = 260.79 cm³/g = 260.79 × 10⁻⁶ m³/kg

Now, we can calculate the ΔU and ΔH:
ΔU = U₂ - U₁ = 2838.6 kJ/kg - 419 kJ/kg
ΔU = 2419.6 kJ/kg

ΔH = H₂ - H₁ = U₂ + P₂v₂ - U₁ - P₁v₁
ΔH = 2838.6 kJ/kg + (800 kPa × 260.79 × 10⁻⁶ m³/kg) - 419 kJ/kg - (100 kPa × 1.044 × 10⁻⁶ m³/kg)
ΔH = 2838.6 kJ/kg + 0.208632 kJ/kg - 419 kJ/kg - 0.1044 kJ/kg
ΔH = 2419.404232 kJ/kg

Therefore, the change in internal energy (ΔU) for the process is approximately 2419.6 kJ/kg, and the change in enthalpy (ΔH) is approximately 2419.404232 kJ/kg.