Question

A rigid, 20-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pressure inside the cooker reaches 150 kPa. Initially, this cooker is filled with water at 175 kPa with a quality of 10 percent. Heat is now added until the quality inside the cooker is 40 percent. Determine the minimum entropy change of the thermal energy reservoir supplying this heat.

A) 120 kJ/K
B) 50 kJ/K
C) 80 kJ/K
D) 30 kJ/K

Answer 1

The minimum entropy change of the thermal energy reservoir supplying heat to the steam cooker is approximately -31 kJ/K.

Step-by-step explanation:

To determine the minimum entropy change of the thermal energy reservoir supplying heat to the steam cooker, we need to calculate the entropy change during the process. We can use the equation:

ΔS = m(Cp)ln(T2/T1)

Where ΔS is the entropy change, m is the mass of the substance, Cp is the specific heat capacity, T2 is the final temperature, and T1 is the initial temperature.

In this case, the thermal energy reservoir is supplying heat to the water inside the steam cooker. The initial temperature T1 is the boiling point temperature at 175 kPa, which is approximately 120°C. The final temperature T2 is the boiling point temperature at 150 kPa, which is approximately 116°C. The specific heat capacity of water is approximately 4.18 J/g°C.

Using these values, we can calculate the minimum entropy change:

ΔS = (20 L)(1000 g/L)(4.18 J/g°C)ln(116°C/120°C)

ΔS ≈ -31 kJ/K

Therefore, the minimum entropy change of the thermal energy reservoir supplying heat to the steam cooker is approximately -31 kJ/K.