mountain climbers know that the atmospheric pressure and the boiling - point of water decrease as elevation…

mountain climbers know that the atmospheric pressure and the boiling - point of water decrease as elevation increases. they approximate linear models that relate them. we measure the pressure in kilopascals (kpa) and the boiling - point of water. at a pressure of 100 kpa the boiling point of water is 100°c and drops by about 3.75°c for each 10 kpa decrease in atmospheric pressure of x kilopascals. g(x)= estimate the boiling point of water (in °c) if the atmospheric pressure is 74 kpa.

mountain climbers know that the atmospheric pressure and the boiling - point of water decrease as elevation increases. they approximate linear models that relate them. we measure the pressure in kilopascals (kpa) and the boiling - point of water. at a pressure of 100 kpa the boiling point of water is 100°c and drops by about 3.75°c for each 10 kpa decrease in atmospheric pressure of x kilopascals. g(x)= estimate the boiling point of water (in °c) if the atmospheric pressure is 74 kpa.

Answer

Explanation:

Step1: Determine the pressure difference

The initial pressure is 100 kPa and the new pressure is 74 kPa. The pressure difference $\Delta P=100 - 74=26$ kPa.

Step2: Calculate the temperature - drop

We know that the boiling - point drops by about 3.75°C for each 10 kPa decrease in pressure. Let $T$ be the temperature drop. Then $T=\frac{26}{10}\times3.75$. $T = 2.6\times3.75=9.75$°C.

Step3: Calculate the new boiling - point

The initial boiling - point at 100 kPa is 100°C. The new boiling - point $B = 100 - T$. $B=100 - 9.75 = 90.25$°C.

Answer:

90.25