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josie411 · Feb 23, 2011, 02:41 PM

hi, I have a question on calculating the GPP in kj energy, from original GPP figures of

4012 g m phytoplankton and energy value of 6 kj g, I thought I needed to convert 4012 g to kj

then divide it by 6 kj g to get the new GPP but am so confused now? Anyone can help please?

jcaron2 · Feb 23, 2011, 09:42 PM

I'm having a hard time understanding your units. For example, I read "4012 g m" as 4012 grams times meters, but I assume you really meant something like 4012 g/m^2? (That grams per square meter). Actually, I thought GPP also was per unit time (like g/m^2/year or something?). No matter. Either way, when expressed in kj energy, I would think you'd have the same units in the denominator, like kj/m^2 or kj/m^2/year or whatever.

If I read your energy value units correctly that should actually be kj/g (kilojoules per gram, not kilojoules times grams), then that means that 1g of phytoplankton equates to an energy of 6 kj. So given that, how much would 2g equate to? 12 kj, right? And 3g would be 18 kj. So now can you tell me how much 4012g would equate to? 4012*6 = 24072 kj, right? So I think that means your GPP in kj energy would be 24072 kj/m^2 (or whatever the heck your units were supposed to be).

Does that answer your question, or did I totally misinterpret what you're asking?

josie411 · Feb 24, 2011, 01:53 AM

Was help, I think I don't understand enough to have put it properly for you, sorry. Let me try again.. GPP in phytoplankton living in ocear area is 4102 g m -2 yr -1 . If you assume energy value of phytoplankton is 6 kj g -1, calc GPP in terms of kj m-2 yr -1?? Sorry cannot get superscript on this site to position figures. Thanks, josie

jcaron2 · Feb 24, 2011, 06:13 AM

Good. In that case, the answer from before should be correct:

$24072 \; kj \cdot m^{-2} \cdot yr^{-1}$

josie411 · Feb 24, 2011, 10:29 AM

Thank you so very much, I will work through your workings so I can make sense of it all.

Josie

jcaron2 · Feb 24, 2011, 11:57 AM

You're welcome.

Here it is written only with math. Hopefully this makes more sense :) :

$4012 \; \frac{g}{m^2 \cdot yr} \; \cdot \; 6 \; \frac{kj}{g}\;=\;24072 \; \frac {kj}{m^2 \cdot yr}$

josie411 · Feb 25, 2011, 02:27 AM

Actually got most of that... which is great, I now have to figure out.. if a woman needs to eat 10000 kj a day, estimate mass of phytoplankton that would meet daily energy needs, then approx what proportion of her own body weight (60 kg) in phytoplankton would she eat in year ?

jcaron2 · Feb 25, 2011, 05:52 AM

Ha! That's a LOT of phytoplankton. More than 10 times her body weight in a year. Yuck! :-)

ebaines · Feb 25, 2011, 12:17 PM

I think what you have is:

$<br /> 10,000 \frac {Kj} {day} \ \times \ \frac 1 6 \frac g {Kj}\ = \ 1667 \frac g {day}<br />$

Josie - can you please define what the term "GPP" means? Thanks.

jcaron2 · Feb 25, 2011, 12:35 PM

It means Gross Primary Production.

Primary production - Wikipedia, the free encyclopedia

AlisonD · Mar 9, 2011, 05:20 AM

Again, Josie, this is a question taken directly from the Open University S104 'Exploring Science' Tutor Marked Assignment 05. If you are having problems you should speak to your tutor because getting the answers from anyone else is considered cheating. I think you will find that the OU anti-plagiarism software is set to scan for questions/answers on sites like this, and you may get yourself in trouble. Be very careful.