Temperature question (help!)

The temperature at sea level is 13 degree celsius.
The temperature at a height of 2000m is 3 degree celsius.

(a) Assuming that the temperature falls at a steady rate as the height increases, find the temperature at a height of 3500m.

(b) A temperatyre below -17 degree celsius is considered dangerous to an aircraft as it will cause severe freezing. Calculate the maximum height that an aircraft can fly up to, before severe freezing takes place.

Please explain your workings clearly and step-by-step. Thanks a lot! :)

The problem is I HAVE the working solutions BUT I do not understand it, so could anyone explain the question and answers in your own way so that I could understand it better? Thanks a lot!


Working solution:

(a) Temperature difference = 3 - 13
= -10

For a height of 3500m, temp difference = (3500/2000) * (-10)
= -17.5
Temp is 13 - 17.5 = -4.5 degree celsius.


(b) Temp. difference = -17 - 13
= - 30

maximum height = (30/10) * 2000
= 6000m


Please help me, thanks! =)

For (a) I can help,
It's not exactly described in the same method, but it works just as good.

Lets look at the difference between 3500 and 2000.
It's 1500.
1500 is 75% of 2000.
Now the decline in temperature for 2000 was 10.
You take 75% from 2000 and also 75% from 10; resulting to 1500 and 7.5.
You subtract 7.5 from 3, and there you have the -4.5 which is the answer.

Quote:

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Originally Posted by Yunie

T

Working solution:

(a) Temperature difference = 3 - 13
= -10 That should be clear!
Now you know that for a climb of 2000m the temperature decreases be 10°
That would make a decrease of (10/2000)° per m of climb.
If you want to climb to 3500m just multiply that by 3500!
3500*(-10/2000)° = (3500/200)*(-10)
For a height of 3500m, temp difference = (3500/2000) * (-10)
= -17.5

Now you know that for a climb of 36500m the temperature would decrease by 17.5°, since the actual temperature at sealevel is 13°, just subtract 17.5° from that one!

Temp is 13 - 17.5 = -4.5 degree celsius.


(b)
The actual Sealevel temperature is 13°, so the temperature difference down to the dangerous temp is -17° -13°, which is -30°

Temp. difference = -17 - 13
= - 30
Now you need to calculate the difference in altitude you would need top get a decrease of 30°. Knowing that for each meter of climb the temperature decreases by (10/2000)° , an the inverse would say that (2000/10) m of climb are needed to get a decrease of 1°.
You need 30° decrease, so it is 30*(200/10) = (30/10)*2000

maximum height = (30/10) * 2000
= 6000m

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BTW: This "dangerous temperature" is nonsense! Severe icing can only happen close to the freezing point (0°),since in temperatures below that, there would be no humidity in the air. It's f****** cold, but not dangerous!

Thanks to the both of you! :) So, the book is lying to me that it is a 'dangerous temperature'...hmm...