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Heat and Thermodynamics question

2016 · 9 Apr · Shift 1 · Q63
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Heat and Thermodynamics question

2016 · 9 Apr · Shift 1 · Q63

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :
  1. A
    8.4 kJ
  2. B
    4.2 kJ
  3. C
    16.7 kJ
  4. D
    167.4 kJ
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of water: m=200 g=0.2 kgm = 200\text{ g} = 0.2\text{ kg}m=200 g=0.2 kg
  • Initial temperature: 40∘C40^\circ\text{C}40∘C
  • Final temperature: 60∘C60^\circ\text{C}60∘C
  • Temperature rise: ΔT=60−40=20 K\Delta T = 60 - 40 = 20\text{ K}ΔT=60−40=20 K
  • Specific heat of water: c=4184 J kg−1K−1c = 4184\,\text{J kg}^{-1}\text{K}^{-1}c=4184J kg−1K−1
  1. Relation for change in internal energy

Ignoring slight expansion of water means work done is negligible, so heat supplied goes almost entirely into increasing internal energy:

ΔU≈mcΔT\Delta U \approx mc\Delta TΔU≈mcΔT

  1. Substitute the values

ΔU=0.2×4184×20\Delta U = 0.2 \times 4184 \times 20ΔU=0.2×4184×20

First compute:

4184×20=836804184 \times 20 = 836804184×20=83680

Then,

0.2×83680=16736 J0.2 \times 83680 = 16736\text{ J}0.2×83680=16736 J

  1. Convert to kJ

16736 J=16.736 kJ16736\text{ J} = 16.736\text{ kJ}16736 J=16.736 kJ

So the change in internal energy is approximately

16.7 kJ\boxed{16.7\text{ kJ}}16.7 kJ​

  1. Option check
  • A: 8.4 kJ8.4\text{ kJ}8.4 kJ ❌
  • B: 4.2 kJ4.2\text{ kJ}4.2 kJ ❌
  • C: 16.7 kJ16.7\text{ kJ}16.7 kJ ✅
  • D: 167.4 kJ167.4\text{ kJ}167.4 kJ ❌

Therefore, the correct option is C.

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