Answer:
The average rate of energy transfer to the cooker is 1.80 kW.
Explanation:
Given that,
Pressure of boiled water = 300 kPa
Mass of water = 3 kg
Time = 30 min
Dryness friction of water = 0.5
Suppose, what is the average rate of energy transfer to the cooker?
We know that,
The specific enthalpy of evaporate at 300 kPa pressure
[tex]h_{f}=561.47\ kJ/kg[/tex]
[tex]h_{fg}=2163.8\ kJ/kg[/tex]
We need to calculate the enthalpy of water at initial state
[tex]h_{1}=h_{f}[/tex]
[tex]h_{1}=561.47\ kJ/kg[/tex]
We need to calculate the enthalpy of water at final state
Using formula of enthalpy
[tex]h_{2}=h_{f}+xh_{fg}[/tex]
Put the value into the formula
[tex]h_{2}=561.47+0.5\times2163.8[/tex]
[tex]h_{2}=1643.37\ kJ/kg[/tex]
We need to calculate the rate of energy transfer to the cooker
Using formula of rate of energy
[tex]Q=\dfrac{m(h_{2}-h_{1})}{t}[/tex]
Put the value into the formula
[tex]Q=\dfrac{3\times(1643.37-561.47)}{30\times60}[/tex]
[tex]Q=1.80\ kW[/tex]
Hence, The average rate of energy transfer to the cooker is 1.80 kW.
