Air thermal

Imagine that a unit of air rises at 3000 meters high, if temperature decreases 6 degrees celcius for every 1000 meter, what will be its temperature at 1400 meters, 2000 meters, 2500 meters and when it reaches the 3000 meter elevation. Starting temperature of air is 35°C.

Result

t1 =  26.6 °C
t2 =  23 °C
t3 =  20 °C
t4 =  17 °C

Solution:

t0=35 t1=t014001000 6=3514001000 6=1335=26.6=26.6Ct_{0}=35 \ ^\circ \text{C} \ \\ t_{1}=t_{0} - \dfrac{ 1400 }{ 1000 } \cdot \ 6=35 - \dfrac{ 1400 }{ 1000 } \cdot \ 6=\dfrac{ 133 }{ 5 }=26.6=26.6 ^\circ \text{C}
t2=t020001000 6=3520001000 6=23=23Ct_{2}=t_{0} - \dfrac{ 2000 }{ 1000 } \cdot \ 6=35 - \dfrac{ 2000 }{ 1000 } \cdot \ 6=23=23 ^\circ \text{C}
t3=t025001000 6=3525001000 6=20=20Ct_{3}=t_{0} - \dfrac{ 2500 }{ 1000 } \cdot \ 6=35 - \dfrac{ 2500 }{ 1000 } \cdot \ 6=20=20 ^\circ \text{C}
t4=t030001000 6=3530001000 6=17=17Ct_{4}=t_{0} - \dfrac{ 3000 }{ 1000 } \cdot \ 6=35 - \dfrac{ 3000 }{ 1000 } \cdot \ 6=17=17 ^\circ \text{C}



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