Plotting lnη vs 1/T for Water and Alcohol Viscosity Measurements

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SUMMARY

This discussion focuses on plotting the graph of lnη versus 1/T for viscosity measurements of water and alcohol. The viscosity values for water at temperatures 293.15 K, 308.15 K, and 323.15 K are 1.002 mPas, 0.726 mPas, and 0.548 mPas, respectively. For alcohol, the corresponding viscosity values are 3.202, 1.254, and 0.705. The equation used for plotting is lnη = (E/R)*(1/T) - lnC, where negative logarithm values for viscosities less than 1 are acceptable for this analysis.

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Homework Statement


Plot the graph of lnη vs 1/T for water (use values from table) and alcohol (values measured previously).
For water:
(Temperature in Kelvin)
T=293.15 η=1.002mPas
T=308.15 η=0.726mPas
T=323.15 η=0.548mPas

For alcohol (measured):
T=293.15 η=3.202
T=308.15 η=1.254
T=323.15 η=0.705

Homework Equations


lnη=(E/R)*(1/T)-lnC

The Attempt at a Solution


I tried just calculating the logarithms of the given values but as expected for values <1 I get negative logarithm values. My question, do I use the negative values as well, or?
 
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