We are asked to calculate the heat of combustion per gram of phenol.

Calculate heat:

$\overline{){\mathbf{Q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{cal}}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}(11.66\frac{\mathrm{kJ}}{\xb0C})(26.37\xb0C-21.36\xb0C)\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{11}\mathbf{.}\mathbf{66}\mathbf{}\frac{\mathbf{kJ}}{\overline{)\mathbf{\xb0}\mathbf{C}}\mathbf{}}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{01}\overline{)\mathbf{\xb0}\mathbf{C}}\mathbf{}\mathbf{)}$

**Q = 58.42 kJ**

A 1.800-g sample of solid phenol (C_{6}H_{5}OH(s)) was burned in a bomb calorimeter, which has a total heat capacity of 11.66 kJ/^{o}C. The temperature of the calorimeter plus its contents increased from 21.36^{o}C to 26.37^{o}C.

You may want to reference (Pages 178 - 183)Section 5.5 while completing this problem.

What is the heat of combustion per gram of phenol?

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