Propan-1-ol Heats Objects Faster than Propan-2-ol. Is this an anomaly?
The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is

$$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -&gt; $a$ CO2(g) + 1/2 $b$ H2O(l)}$$

The corresponding standard enthalpy of combustion is

$$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$

This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$.

The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align}
\Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&amp;=-302.6\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&amp;=-318.1\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{H2O, l})&amp;=-285.8\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{CO2, g})&amp;=-393.5\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.