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Aloiza [94]
3 years ago
12

Archeologists can determine the age of artifacts made of wood or bone by measuring the amount of the radioactive isotope 14C pre

sent in the object. The amount of isotope decreases in a first-order process. If 15.5% of the original amount of 14C is present in a wooden tool at the time of analysis, what is the age of the tool? The half-life of 14C is 5730 yr.
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer: The age of the tool is 15539 years

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant = 100

a - x = amount left after decay process = \frac{15.5}{100}\times 100=15.5  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.  

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{5730yr}=0.00012yr^{-1}

b) for 15.5 % of original amount

t=\frac{2.303}{0.00012}\log\frac{100}{15.5}

t=15539years

Thus age of the tool is 15539 years

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Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

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Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

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Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

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Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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