Answer:
The atmospheric pressure is 0.843 bar.
Explanation:
First, we calculate how much thermal energy (heat) was transferred to the water:
[tex]\mbox{Heat = Power * time * efficiency}\\\\\mbox{Heat}=2 kW*30min*60\frac{s}{min}*0.75=2700 kJ[/tex]
Knowing that all heat (Q) was used in making 1.19 kg of water boil (liquid to gas), we can find what is the latent heat ([tex]L_{vaporization}[/tex]) of this change of state (vaporization):
[tex]L = \frac {Q}{m}=\frac {2700kJ}{1.19kg}=2268.9{J}{g}[/tex]
Looking up this value in a Water Heat of Vaporization Calculator, we find that it corresponds to a temperature of 94.9°C.
Knowing the temperature at which the water boils, we have to find the vapor pressure (the same as the latent heat according to temperature, it is a value which can be found in a table) at that temperature, which would be the atmospheric pressure of the location.
The vapor pressure of water at 94.9°C is 0.843 bar, i.e. the atmospheric pressure is 0.843 bar.
