Final answer:
The work associated with the isothermal process of water is 11000 kJ, and the entropy generated is 25996.57 J/K.
Step-by-step explanation:
To determine the work associated with the process and the entropy generated during the process when 4.5 kg of water at 1 MPa, 150°C undergoes a reversible isothermal process with 11000 kJ of heat provided, we can use the first and second laws of thermodynamics. Since it is an isothermal process, the change in internal energy (dU) will be zero. Therefore, the work done (W) by the system will be equal to the heat transferred into the system (Q).
W = Q = 11000 kJ
To find the entropy change (ΔS), we can use the formula:
ΔS = Q / T
However, we need to convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius value.
T(K) = 150°C + 273.15 = 423.15 K
Substituting the values:
ΔS = 11000 kJ / 423.15 K
To get the units of entropy in J/K, we also need to convert kJ to J by multiplying by 1000.
ΔS = (11000 kJ * 1000) / 423.15 K = 25996.57 J/K
Therefore, the work associated with the process is 11000 kJ, and the entropy generated during the process is 25996.57 J/K.