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Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q50
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  5. /2022 · 25 Jun · Shift 2 · Q50

Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q50

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A copper block of mass 5.0 kg is heated to a temperature of 500 ∘^\circ∘ C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper : 0.39 J g −-− 1 ∘^\circ∘ C −-− 1 and latent heat of fusion of water : 335 J g −-− 1]
  1. A
    1.5 kg
  2. B
    5.8 kg
  3. C
    2.9 kg
  4. D
    3.8 kg
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of copper block: mc=5.0 kg=5000 gm_c = 5.0\,\text{kg} = 5000\,\text{g}mc​=5.0kg=5000g
  • Initial temperature of copper: 500∘C500^\circ\text{C}500∘C
  • Final temperature when placed on ice: 0∘C0^\circ\text{C}0∘C
  • Specific heat of copper: c=0.39 J g−1 ∘C−1c = 0.39\,\text{J g}^{-1}\,{}^\circ\text{C}^{-1}c=0.39J g−1∘C−1
  • Latent heat of fusion of ice: L=335 J g−1L = 335\,\text{J g}^{-1}L=335J g−1

Since the ice block is large, the copper cools down to 0∘C0^\circ\text{C}0∘C and all the heat lost by copper is used to melt ice.

  1. Heat lost by copper
Q=mccΔTQ = m_c c \Delta TQ=mc​cΔT

Here,

ΔT=500−0=500∘C\Delta T = 500 - 0 = 500^\circ\text{C}ΔT=500−0=500∘C

So,

Q=5000×0.39×500Q = 5000 \times 0.39 \times 500Q=5000×0.39×500 Q=975000 JQ = 975000\,\text{J}Q=975000J
  1. Ice melted

If mmm grams of ice melts, then

Q=mLQ = mLQ=mL m=QL=975000335m = \frac{Q}{L} = \frac{975000}{335}m=LQ​=335975000​ m≈2910.45 gm \approx 2910.45\,\text{g}m≈2910.45g m≈2.91 kgm \approx 2.91\,\text{kg}m≈2.91kg
  1. Match with options

The closest option is:

C: 2.9 kg\boxed{\text{C: } 2.9\,\text{kg}}C: 2.9kg​
  1. Comparison with stored answer

Stored correct answer is C, which matches the derived answer.

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