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AIIMS 2016
Exam
AIIMS
Year
2016
Subject
Chemistry
Chapter
Thermodynamics (C)

AIIMS 2016 · Chemistry · Thermodynamics (C)

The heat liberated when 1.89 ~g of benzoic acid is burnt in a bomb...

The heat liberated when 1.89 g1.89 \mathrm{~g} of benzoic acid is burnt in a bomb calorimeter at 25∘C25^{\circ} \mathrm{C} and it increases the temperature of 18.94 kg18.94 \mathrm{~kg} of water by 0.632∘C0.632^{\circ} \mathrm{C}. If the specific heat of water at 25∘C25^{\circ} \mathrm{C} is 0.998cal/g0.998 \mathrm{cal} / \mathrm{g} - deg\mathrm{deg}, the value of the heat of combustion of benzoic acid is

Options

  1. A881.1kcal881.1 \mathrm{kcal}
  2. B771.12kcal771.12 \mathrm{kcal}
  3. C981.1kcal981.1 \mathrm{kcal}
  4. D871.2kcal871.2 \mathrm{kcal}

Correct answer

B. 771.12kcal771.12 \mathrm{kcal}

Step-by-step solution

Given : Weight of benzoic acid =1.89 g=1.89 \mathrm{~g}; Temperature of bomb calorimeter =25∘C=298 K=25^{\circ} \mathrm{C}=298 \mathrm{~K}; Mass of water (m)=18.94 kg=18940 g(m)=18.94 \mathrm{~kg}=18940 \mathrm{~g}; Increase in temperature (Δt)=0.632∘C(\Delta t)=0.632^{\circ} \mathrm{C} and specific heat of water (s)=0.998cal/g(s)=0.998 \mathrm{cal} / \mathrm{g}-deg. We know that heat gained by water or heat liberated by benzoic acid (Q)=msΔt(Q)=m s \Delta t =11946.14×1221.89=771126.5cal=771.12kcal=\frac{11946.14 \times 122}{1.89}=771126.5 \mathrm{cal}=771.12 \mathrm{kcal}
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