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

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The discussion focuses on plotting the graph of lnη versus 1/T for water and alcohol using provided viscosity data. For water, viscosity values at three temperatures in Kelvin are given, while alcohol viscosity values were measured previously at the same temperatures. The equation lnη = (E/R)(1/T) - lnC is referenced for the calculations. A participant expresses confusion about handling negative logarithm values for viscosities less than one. It is confirmed that negative values should be used in the calculations.
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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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