Question

Water is boiled in a pan covered with a poorly fitting lid at a specified location. Heat is supplied to the pan by a 2-kW resistance heater. The amount of water in the pan is observed to decrease by 1.19 kg in 30 minutes. If it is estimated that 75 percent of electricity consumed by the heater is transferred to the water as heat, determine the local atmospheric pressure in that location.

Answer 1

The atmospheric pressure is 0.843 bar.

Step-by-step explanation:

First, we calculate how much thermal energy (heat) was transferred to the water:

`\mbox{Heat = Power * time * efficiency}\\\\\mbox{Heat}=2 kW*30min*60(s)/(min)*0.75=2700 kJ`

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 (
`L_(vaporization)`) of this change of state (vaporization):

`L = \frac {Q}{m}=\frac {2700kJ}{1.19kg}=2268.9{J}{g}`

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.