Imagine

Imagine that a unit of air at a temperature of 25°C rises up a mountain range that is 3,000 meters high on the windward side and which descends to 1,200 meters on the leeward side, assuming that the air will remain dry what will its temperature when it crosses the top of the range and what will it be when it descends to the base on the lee side? (temperature decreases 6 degrees Celcius for every 1000 meter)

Result

t1 =  7 °C
t2 =  17.8 °C

Solution:

t1=256 30001000=7=7Ct_{ 1 } = 25 - 6 \cdot \ \dfrac{ 3000 }{ 1000 } = 7 = 7 ^\circ C
t2=256 12001000=895=17.8=17.8Ct_{ 2 } = 25 - 6 \cdot \ \dfrac{ 1200 }{ 1000 } = \dfrac{ 89 }{ 5 } = 17.8 = 17.8 ^\circ C



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