Saladsamurai
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I do not understand how they went from [itex]\ln T_c-\lnT_H=\theta_c-\theta_H[/itex]
to
[itex]\theta=\ln T+C[/itex]
Why did the subscripts drop out? What just happened?
KELVIN scale-Can't make sense of the last step of this solution
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I do not understand how they went from [itex]\ln T_c-\lnT_H=\theta_c-\theta_H[/itex]
to
[itex]\theta=\ln T+C[/itex]
Why did the subscripts drop out? What just happened?
Discussion
physicsnoob93
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exp means e^thethac if that is thetha
so you get e^(thethac-thethah) is Tc/Th.
If you ln both sides, you'll get ln(Tc/Th) = lnTc-lnTh and on the other side lne^(thethac-thethah) = (thethac-thethah)lne. lne is one.
Thus
lnTc-lnTh = thethac-thethah
so you get e^(thethac-thethah) is Tc/Th.
If you ln both sides, you'll get ln(Tc/Th) = lnTc-lnTh and on the other side lne^(thethac-thethah) = (thethac-thethah)lne. lne is one.
Thus
lnTc-lnTh = thethac-thethah
Science Advisor
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"C" and "H" just represent different measurements of the temperature. Since you take TH to be a fixed temperature, that just leaves one temperature- you don't need the "C" to distinguish it form "H".
Saladsamurai
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I am sorry Halls, I don't quite follow. What do you mean I take TH to be fixed?