Answer:
[tex]P_{atm} = 87.5\,kPa[/tex]
Explanation:
The heat required to boil the water in the pan is:
[tex]Q = \eta_{e}\cdot \dot W_{e} \cdot \Delta t[/tex]
[tex]Q = 0.75 \cdot (2\,kW)\cdot (30\, min)\cdot (\frac{60\,sec}{1\, min} )[/tex]
[tex]Q = 2700\,kJ[/tex]
Since the pan is accompained with a poorly fitting lid, the heating process is isobaric and change on specific enthalpy is obtained by following expression:
[tex]\Delta h = \frac{Q}{\Delta m}[/tex]
[tex]\Delta h = \frac{2700\,kJ}{1.19\,kg}[/tex]
[tex]\Delta h = 2268.908\,\frac{kJ}{kg}[/tex]
Then, the local atmospheric pressure can be estimated by looking for the saturation pressure related to the change on specific enthalpy at property tables for saturated water:
[tex]P_{atm} = 87.5\,kPa[/tex]
