Question

A 0.0604−mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 1.41 × 102 g H2O. Given that the fuel value is 4.53 × 10−2 in nutritional Cal when the temperature of the water is increased by 2.21°C, calculate the fuel value in kJ.

Answer 1

Answer on Question # 72786 - Chemistry - General Chemistry

A 0.0604-mol sample of a nutrient substance, with a formula weight of $114\mathrm{g/mol}$, is burned in a bomb calorimeter containing $1.41 \times 10^{2}\mathrm{g}$ H2O. Given that the fuel value is $4.53 \times 10^{-2}$ in nutritional Cal when the temperature of the water is increased by $2.21^{\circ}\mathrm{C}$, calculate the fuel value in kJ.

Solution:

The only required data here is the fuel value in nutritional calories, which is $4.53 \times 10^{-2}$ Cal. Using the conversion $1\mathrm{Cal} = 4.184\mathrm{kJ}$, we can set up a simple conversion:

$$4.53 \times 10^{-2} \mathrm{Cal} \ (4.184\ \mathrm{kJ} / 1\ \mathrm{Cal}) = 0.190\ \mathrm{kJ}.$$

Answer: 0.190 kJ.

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