Question

What is the boiling point of water at an elevation of 1.55×104 ft ?

Answer 1

Answer:

85.0°C

Explanation:

The boiling point of water at sea level is 100°C and at this level, the pressure is 1 atm or 760 mmHg. The pressure decreases at a linear function when the high increases. At every 1,000 ft the pressure decreases 19.8 mmHg, thus:

1,000 ft ----------------- 19.8 mmHg

1.55x10⁴ft ------------- x

By a simple direct three rule:

1000x = 306,900

x = 306.9 mmHg

The boiling point of water decreases 0.05°C for each 1 mmHg pressure decreases, so:

1 mmHg ------- 0.05°C

306.9 mmHg --- y

By a simple direct three rule:

y = 15.345°C

So, the temperature will be 100°C - 15.345°C = 84.655 °C = 85.0°C

Answer 2

The solution for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease 
= 308.14 x 0.05 C = 15.407 deg C 
deduct this amount to 100
100 – 15.407 = 84.593 C 
ANSWER: 85 deg C (rounded to 2 significant figures)