GPH Lab 31
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GPH Lab 3
In this lab students will practice converting units of measure between the metric system and the
“American” system; practice subtracting decimal numbers; and use the average lapse rate.
Converting measurements of distance
1.
65 miles = _____kilometers (for an approximate result, multiply the length value by 1.609).
Show your work for full credit.
65 * 1.609 = 104.585
2.
25 kilometers = ___________miles (for an approximate result, divide the length value by
1.609). Show your work for full credit.
25 / 1.609 = 15.538
3.
2 inches = _________centimeters (multiply the length value by 2.54). Show your work for full
credit.
2 * 2.54 = 5.08
Subtracting decimals using Figure 3-3 (page 59 in the textbook)
4.
What is the difference in percentage of nitrogen in the atmosphere to the percentage of
carbon dioxide? Show your work for full credit.
According to the Figure 3-3 in the textbook:
Nitrogen - 78.08% in the atmosphere
Carbon Dioxide – 0.04% in the atmosphere
78.08% - 0.04% = 78.04%
There is 78.04 % more nitrogen in the atmosphere than Carbon Dioxide.
5.
What is the difference in percentage of oxygen in the atmosphere to the percentage of
argon? Show your work for full credit.
According to the Figure 3-3 in the textbook:
Oxygen – 20.95% in the atmosphere
Argon – 0.93% in the atmosphere
20.95% - 0.93% = 20.02%
There is 20.02% more oxygen in the atmosphere than Argon.
Converting measurements of temperature
6.
20°C = _____°F To convert from Celsius to Fahrenheit: (°C×1.8)+32°=°F
Show your work for full credit.
20*1.8= 36+32= 68°F
7.
120°F = _____°C To convert from Fahrenheit to Celsius: (°F −32°)÷1.8=°C
Show your work for full credit.
120-32=88 /1.8 = 48.88 or 49°C
Calculating temperature using the average lapse rate
8.
If the temperature of the air is 15°C at sea level, use the average lapse rate (6.5°C per 1000
meters) to calculate the expected temperature at the top of a 5000-meter-high mountain:
_____°C. Show your work for full credit.
The temperature will decrease with increasing altitude at a rate of 6.5°C per 1000 meters
elevation change.
5000 meters – 0 meters (sea level is 0 elevation) = 5000 meters difference in elevation
6.5°C * 5 (lapses) = 32.5°C total change in temperature
15°C - 32.5°C = -17.5°C at 5000 meters altitude
The temperature at the top of a 5000-meter-high mountain would be –17.5°C.
9.
If the air temperature outside a jet airplane flying at 36,000 feet is −30°F, use the average
lapse rate (3.6°F per 1000 feet) to estimate the temperature at sea level below: _____°F.
Show your work for full credit.
The temperature will increase with decreasing altitude at a rate of 3.6°F per 1000 feet elevation
change.
3.6 * 36 (lapses) = 129.
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