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MMLU-Pro / 9271 / Benzoic acid, 1.35 mathrm{~g}, is reacted with oxygen in a constant volume…

Problem

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question

Benzoic acid, $1.35 \mathrm{~g}$, is reacted with oxygen in a constant volume calorimeter to form H2O(l)\mathrm{H}_2 \mathrm{O}(l) and CO2(g)\mathrm{CO}_2(g) at $298 \mathrm{~K}$. The mass of the water in the inner bath is 1.55×1.55 \times $10^3 \mathrm{~g}$. The temperature of the calorimeter and its contents rises $2.76 \mathrm{~K}$ as a result of this reaction. Calculate the calorimeter constant.
Plain-text mathematical notation (without MathML)
Benzoic acid, $1.35 \mathrm{~g}$, is reacted with oxygen in a constant volume calorimeter to form H₂O(l) and CO₂(g) at $298 \mathrm{~K}$. The mass of the water in the inner bath is 1.55× $10^3 \mathrm{~g}$. The temperature of the calorimeter and its contents rises $2.76 \mathrm{~K}$ as a result of this reaction. Calculate the calorimeter constant.
Original LaTeX notation
Benzoic acid, $1.35 \mathrm{~g}$, is reacted with oxygen in a constant volume calorimeter to form $\mathrm{H}_2 \mathrm{O}(l)$ and $\mathrm{CO}_2(g)$ at $298 \mathrm{~K}$. The mass of the water in the inner bath is $1.55 \times$ $10^3 \mathrm{~g}$. The temperature of the calorimeter and its contents rises $2.76 \mathrm{~K}$ as a result of this reaction. Calculate the calorimeter constant.

category

physics

options

  1. 5.98 x 10^3 J°C^-1
  2. 7.05 x 10^3 J°C^-1
  3. 6.64 $10^3 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$
  4. 4.50 x 10^3 J°C^-1
  5. 10.12 x 10^3 J°C^-1
  6. 5.23 $10^3 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$
  7. 8.42 $10^3 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$
  8. 7.89 $10^3 \mathrm{~J}^{\circ} \mathrm{C}^{-1}$
  9. 3.30 x 10^3 J°C^-1
  10. 9.75 x 10^3 J°C^-1

src

scibench-thermo

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