Explanation of problem please

K

Kezzer

New member
Joined
Sep 23, 2019
Messages
5
I would be extremely grateful if someone can help me with this problem. I understand part of it, but not the final section.

Residence time equals: 1.5x1016kg of water (H2O) divided by 5.05x1017kg of H2O

What is the residence time of H2O in the atmosphere to the nearest day?

So for the equation I get 2.97, but the answer gives the following workings out and I don't understand where all the other bits come from:

(2.97x10-2y) 1/y-1
= 2.97 x10-2y
=0.0297y x 365.4 days y-1
= 11 days

Many thanks for your help.
 
Kezzer said:
Year?
Click to expand...
Are you asking me? I am not familiar with the formula. Please look it up.
Also, double check your answer - are you sure it's 2.97 (years)?
So, assuming the correct answer is 0.0297 years, how do we convert it into days?
1 year is 365.4 days
0.0297 years is X days
X = ?
 
One thing you need to do is to show us the entire problem, rather than drop us into the middle of your work. I imagine the problem gave some units, which we need to know; it will also clarify what you are required to find.

Then, as was mentioned, check your work; you made a mistake with the scientific notation.
 
Hi, the info I have is as follows:

Dividing the mass of a substance in a reservoir by its rate of transfer through that reservoir gives us its residence time:

Residence time equals: mass of substance in reservoir divided by rate at which substance enters (and/or leaves) reservoir,

or the average time that a substance spends in a reservoir.

The atmosphere contains 1.5 x 1016kg of water (H2O) mostly as water vapour, and precipitation and evaporation are balanced, transferring 5.05 x 1017kg of H2O per year between the atmosphere and the oceans/land.

What is the residence time of H2O in the atmosphere to the nearest day?

Many thanks for your input.
 
Kezzer said:
The atmosphere contains 1.5 x 1016kg of water (H2O) mostly as water vapour, and precipitation and evaporation are balanced, transferring 5.05 x 1017kg of H2O per year between the atmosphere and the oceans/land.

What is the residence time of H2O in the atmosphere to the nearest day?
Click to expand...

The bits I've bolded are what we need. You calculated the residence time in years, since you divided a mass in kg by a rate in kg/year. Then you had to convert that to days.

Now, let's get the calculation fixed: what is 1.5 x 1016kg divided by 5.05 x 1017? It will help if you show the steps you took.
 
I have cracked it: never do things when you are tired!

So 1.5 x 1016 divided by 5.05 x 1017 = 0.02970297029 or 10-2y

0.0297y x 365.4 days y-1 = 11 days

Or at least I think I have! Thanks everyone for your input.
 
Kezzer said:
I have cracked it: never do things when you are tired!

So 1.5 x 1016 divided by 5.05 x 1017 = 0.02970297029 or 10-2y

0.0297y x 365.4 days y-1 = 11 days

Or at least I think I have! Thanks everyone for your input.
Click to expand...
Yes, that looks good.

I assume what you meant there was "or 2.97 x 10-2y".

Presumably you have no trouble with the notation you and your book are using, where "365.4 days y-1" means 365.4 days per year, as a conversion factor. I haven't yet figured out what was meant in your OP, by "(2.97x10-2y) 1/y-1", where the last bit seems out of place.
 
Apologies for the delay in responding: I have major problems with my phone line and internet connection. Thank you for correcting me on the "or 2.97 x 10-2yr". Yes I understand the y_1. I haven't figured out the 1/y-1 either, but in the line below that it gives the figures without the brackets and without that 1/y-1, so I can't understand the point of it. Many thanks for your help.
 
You must log in or register to reply here.
Top