What will most likely happen when a warmer air mass exists above a polar ice sheet?

2 answers:

8 0

I believe it is "The temperature of the ice sheet will be lowered" because it will most likely lower just a bit due to the heat. It's not going to fully melt but it will lower just a bit.
Plus I like your profile pic.

adelina 88 [10]2 years ago

7 0

Answer: Option (b) is the correct answer.

Explanation:

When warm air will rise above a polar ice sheet then it means due to high kinetic energy between the warm air molecules they will collide rapidly and frequently from one place to another.

As a result, warm air molecules will expand at a faster rate. As polar ice sheet is colder in nature therefore, it will absorb the heat from warm air.

Hence, the polar ice sheet will start to melt as absorption of heat will generate kinetic energy between the molecules of polar ice sheet.

Thus, we can conclude that heat will be transferred from the air mass to the ice.

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The half-life for the reaction below was determined to be 2.14 × 10^4 s at 800 K. The value of the half-life is independent of t

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Answer:

0.0907 s

Explanation:

This an Arrhenius equation problem, so you relate the half-life with the kinetic constant of the reaction in order to calcule the same thermodynamic parameters at another temperature.

To calcule the kinetic constant of the reaction you need to know the order of it, look closely to the sentence "The value of the half-life is independent of the inital concentration of N2O present." the only order independent from the initial concentration of reagents is first order, so you can calculate K at 800 K, using:

$k(800 K)=\frac{ln(2)}{t_{1/2}}= \frac{ln(2)}{2.14 * 10^{4} s}}=3.239*10^{-5}s^{-1}}$

Now you can use Arrhenius equation to calcule K at 1150.66 K

$ln(\frac{k1}{k2} )=-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} )$

$k2= k1*exp(-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} ))=3.239*10^{-5}s^{-1}*exp(-\frac{270 000 J/mol }{8.314 J/mol *k }(\frac{1}{1150.66K} - \frac{1}{800K} ))=7.639 s^{-1}$