turnover ere is a total of 10.4 kg C/m\' in the grassland soil pictured in Figur
ID: 119495 • Letter: T
Question
turnover ere is a total of 10.4 kg C/m' in the grassland soil pictured in Figure 5.19. Calculate the of the soil organic carbon with respect to each of the three components of soil respiration and with respect to total soil respiration. Then, calculate the overall turnover time with respect to total inputs to the soil. How do these compare? Is the overall turnover time controlled by the pools with rapid or slow turnover? Soil respiration Aboveground itterfall 0.004 0.005 0.419 0.16 0.26 Undecomposed litter e Turnover in 10s of years Roots 0.80 kg C/m2 turnover 0.004 Fulvic acids 2.3 kg C/m Turnover in 100s of years 0,003X0.003 Humins 3.8 kg C/m2 Humic acids 3.8 kg C/m2 Turnover in 1000s of years Permanent accumulations in the lower profile FIGURE 5.19 Turnover of detritus and soil organic fractions in a grassland soil, in units of kgC m2 yr1, Note that mean residence time can be calculated for each fraction from measurements of the quantity in the soil and the annual production or loss (respiration) from that fraction. Source: From Schlesinger (1977) 20 Copyright© 2013 Elsevier Inc. All rights reserved.Explanation / Answer
4)Answer:
Turn over undecomposed litter = (0.5+0.26)-(0.41+0.004+0.006)=0.34 Kg C/m2
Turn over Fulvic acid = 2.3+0.004-0.004 = 2.3 Kg C/m2
Turnover of Humin and Humic acid =3.8+3.8+0.003+0.003-0.005= 7.601 Kg C/m2
Overall turnover = (0.5+2.3+3.8+3.8+0.8)+0.16 – 0.419= 10.941 Kg C/m2
Total input= 0.26+0.16 = 0.42
For undecomposed litter, input rate= 0.42/10s of year in 0.5 Kg C/m2.
Then, residence time= 0.5/input rate = (0.5/0.42)* 10s of year=11.9s of year.
For fulvic acid, residence time = 2.3/input rate = (2.3/0.004)*100s of year = 57500s of year
For humin & humic acid, residence time = 7.6/input rate = (7.6/0.006)*1000s of year= 12,66666.67s of year
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