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Heat and Thermodynamics question

2022 · 24 Jun · Shift 2 · Q58
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  5. /2022 · 24 Jun · Shift 2 · Q58

Heat and Thermodynamics question

2022 · 24 Jun · Shift 2 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms −-− 1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron = 0.42 Jg −-− 1 ∘^\circ∘ C −-− 1]
  1. A
    675 ∘^\circ∘ C
  2. B
    1600 ∘^\circ∘ C
  3. C
    16.07 ∘^\circ∘ C
  4. D
    6.75 ∘^\circ∘ C
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of hammer: mh=1.5 kgm_h = 1.5\,\text{kg}mh​=1.5kg
  • Velocity of hammer: v=60 m s−1v = 60\,\text{m s}^{-1}v=60m s−1
  • Mass of iron nail: m=100 gm = 100\,\text{g}m=100g
  • Specific heat of iron: c=0.42 J g−1 ∘C−1c = 0.42\,\text{J g}^{-1}\,^{\circ}\text{C}^{-1}c=0.42J g−1∘C−1
  • One fourth of hammer's kinetic energy heats the nail.
  1. Calculate kinetic energy of the hammer
KE=12mhv2KE = \frac{1}{2} m_h v^2KE=21​mh​v2 KE=12(1.5)(60)2KE = \frac{1}{2}(1.5)(60)^2KE=21​(1.5)(60)2 KE=0.75×3600=2700 JKE = 0.75 \times 3600 = 2700\,\text{J}KE=0.75×3600=2700J
  1. Energy transferred to the nail

Since only one fourth of this energy goes into heating the nail,

Q=14×2700=675 JQ = \frac{1}{4} \times 2700 = 675\,\text{J}Q=41​×2700=675J
  1. Use heat formula
Q=mcΔTQ = mc\Delta TQ=mcΔT

So,

ΔT=Qmc\Delta T = \frac{Q}{mc}ΔT=mcQ​

Substitute the values:

ΔT=675(100)(0.42)\Delta T = \frac{675}{(100)(0.42)}ΔT=(100)(0.42)675​ ΔT=67542=16.07∘C\Delta T = \frac{675}{42} = 16.07^{\circ}\text{C}ΔT=42675​=16.07∘C
  1. Match with options

The rise in temperature is:

16.07∘C\boxed{16.07^{\circ}\text{C}}16.07∘C​

So the correct option is C.

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