Chapter 20, Problem 2PEB

The average rate of chemical weathering of carbonate rock in an urban area is 2.1 $\times$ 10^–3 mm/yr. The depth of lettering of the tombstones shown in the following table was measured in 2010:

Date on Tombstone	Age (yr)
1910	100
1921	89
1948	62
1958	52
1975	35

How much has the depth of the lettering been reduced on each tombstone?

To determine

The depth of the lettering been reduced on each tombstone if the average rate of chemical weathering of carbonate rock in an urban area is $2.1\times {10}^{-3}\text{ mm/yr}$.

Answer to Problem 2PEB

Solution:

$2.1\times {10}^{-1}\text{ mm,}1.86\times {10}^{-1}\text{ mm,}1.30\times {10}^{-1}\text{ mm,}1.09\times {10}^{-1}\text{ mm and }7.35\times {10}^{-2}\text{ mm}$

Explanation of Solution

Given data:

The average rate of chemical weathering of carbonate rock in an urban area is $2.1\times {10}^{-3}\text{ mm/yr}$.

The depth of lettering of the tomstones shown in the following table was measured by 2010.

$\begin{array}{ll}\text{Date on Tombstone}\hfill & \text{Age}\left(\text{yr}\right)\hfill \\ 1910\hfill & 100\hfill \\ 1921\hfill & 89\hfill \\ 1948\hfill & 62\hfill \\ 1958\hfill & 52\hfill \\ 1975\hfill & 35\hfill \end{array}$

Formula used:

Write the weathering rate is divided by the surface reduction by the age of the tombstone.

$\text{Rate}=\frac{\text{Reduction}}{\text{age}}$

Explanation:

Recall that the weathering rate is divided by the surface reduction by the age of the tombstone.

$\text{Rate}=\frac{\text{Reduction}}{\text{age}}$

Rearranged:

$\text{Reduction}=\text{Rate}\left(\text{age}\right)$ …… (1)

For date of tombstone 1920 and age $100\text{ yr}$:

Substitute $2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}$ for rate and $100\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\left(\text{Reduction}\right)}_{1910}=2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}\left(100\text{ yr}\right)\\ =2.1\times {10}^{-1}\text{ mm}\end{array}$

For date of tombstone 1921 and age $89\text{ yr}$:

Substitute $2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}$ for rate and $89\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\left(\text{Reduction}\right)}_{1921}=2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}\left(89\text{ yr}\right)\\ =1.86\times {10}^{-1}\text{ mm}\end{array}$

For date of tombstone 1948 and age $62\text{ yr}$:

Substitute $2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}$ for rate and $62\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\left(\text{Reduction}\right)}_{1948}=2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}\left(62\text{ yr}\right)\\ =1.30\times {10}^{-1}\text{ mm}\end{array}$

For date of tombstone 1958 and age $52\text{ yr}$:

Substitute $2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}$ for rate and $52\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\left(\text{Reduction}\right)}_{1958}=2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}\left(52\text{ yr}\right)\\ =1.09\times {10}^{-1}\text{ mm}\end{array}$

For date of tombstone 1975 and age $35\text{ yr}$:

Substitute $2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}$ for rate and $35\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\left(\text{Reduction}\right)}_{1975}=2.1\times {10}^{-3}\frac{\text{mm}}{\text{yr}}\left(35\text{ yr}\right)\\ =7.35\times {10}^{-2}\text{ mm}\end{array}$

Conclusion:

The depth of the lettering been reduced on each tombstones are $2.1\times {10}^{-1}\text{ mm,}1.86\times {10}^{-1}\text{ mm,}1.30\times {10}^{-1}\text{ mm,}1.09\times {10}^{-1}\text{ mm and }7.35\times {10}^{-2}\text{ mm}$, respectively.