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Heat and Thermodynamics question

2024 · 27 Jan · Shift 1 · Q70
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Heat and Thermodynamics question

2024 · 27 Jan · Shift 1 · Q70

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
0.08 kg0.08 \mathrm{~kg}0.08 kg air is heated at constant volume through 5∘C5^{\circ} \mathrm{C}5∘C. The specific heat of air at constant volume is 0.17 kcal/kg∘C0.17 \mathrm{~kcal} / \mathrm{kg}^{\circ} \mathrm{C}0.17 kcal/kg∘C and J=4.18\mathrm{J}=4.18J=4.18 joule/ cal\mathrm{~cal} cal. The change in its internal energy is approximately.
  1. A
    318 J
  2. B
    298 J
  3. C
    284 J
  4. D
    142 J
View written solutionFree

Correct answer: C

  1. Use the relation for change in internal energy at constant volume

At constant volume,

ΔU=mcvΔT\Delta U = m c_v \Delta TΔU=mcv​ΔT

where:

  • m=0.08 kgm = 0.08\,\text{kg}m=0.08kg
  • cv=0.17 kcal/(kg∘C)c_v = 0.17\,\text{kcal/(kg}^\circ\text{C)}cv​=0.17kcal/(kg∘C)
  • ΔT=5∘C\Delta T = 5^\circ\text{C}ΔT=5∘C
  1. Compute heat in kcal
ΔU=0.08×0.17×5\Delta U = 0.08 \times 0.17 \times 5ΔU=0.08×0.17×5 ΔU=0.068 kcal\Delta U = 0.068\,\text{kcal}ΔU=0.068kcal
  1. Convert kcal to joules

Given:

1 cal=4.18 J1\,\text{cal} = 4.18\,\text{J}1cal=4.18J

So,

1 kcal=1000×4.18=4180 J1\,\text{kcal} = 1000 \times 4.18 = 4180\,\text{J}1kcal=1000×4.18=4180J

Hence,

ΔU=0.068×4180\Delta U = 0.068 \times 4180ΔU=0.068×4180 ΔU=284.24 J\Delta U = 284.24\,\text{J}ΔU=284.24J
  1. Approximate value
ΔU≈284 J\Delta U \approx 284\,\text{J}ΔU≈284J
  1. Check options
  • A: 318 J318\,\text{J}318J
  • B: 298 J298\,\text{J}298J
  • C: 284 J284\,\text{J}284J ✅
  • D: 142 J142\,\text{J}142J

Therefore, the correct option is C.

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