Answer:
84.75°C is the boiling point of water at an elevation of 7000 meter.
Explanation:
Rate of change of pressure = 19.8 mmHg/1000 ft
1 foot = [tex]\frac{1}{3.28} [/tex] meter
[tex]19.8 mmHg/1000 ft=\frac{19.8}{1000}\times 3.28 mmHg/m=0.065 mmHg[/tex]
Pressure change for every 1 m = 0.065 mmHg × 1= 0.065 mmHg
Elevation Pressure
0 m 760 mmHg
1000 m 695 mmHg
2000 m 630 mmHg
Pressure drop at the elevation of 7000 m: [tex]7000\times 0.065 mmHg=455 mmHg[/tex]
Pressure at 7000 m = 760 mmHg - 455mmHg = 305 mmHg
The boiling point of water decreases 0.05°C for every 1 mmHg drop in atmospheric pressure.
At 7000 meter elevation the boiling of water will be :
[tex]0.05^oC\times 305=15.25^oC[/tex]
Boiling point of water at 7000 meter elevation :
[tex]100.0^oC-15.25^oC =84.75^oC[/tex]
84.75°C is the boiling point of water at an elevation of 7000 meter.