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JEE Main 2023 (29 Jan, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2023 (29 Jan, Shift 2)

Q19JEE Main 2023 · 29 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Heat energy of 184 kJ is given to ice of mass 600 g at −12∘C-12^{\circ}\text{C}. Specific heat of ice is 2222.3 J kg−1 ∘C−12222.3\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1} and latent heat of ice is 336 kJ kg−1\text{kJ kg}^{-1}

A. Final temperature of system will be 0∘^{\circ} C.

B. Final temperature of the system will be greater than 0∘^{\circ}C.

C. The final system will have a mixture of ice and water in the ratio of 5: 1.

D. The final system will have a mixture of ice and water in the ratio of 1: 5.

E. The final system will have water only.

Choose the correct answer from the options given below:

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Answer: (d)

  1. Ice −12→0∘C-12 \to 0^{\circ}\text{C}: 0.6×2222.3×12≈16 kJ0.6 \times 2222.3 \times 12 \approx 16\text{ kJ}.
  2. Heat left =184−16=168 kJ= 184 - 16 = 168\text{ kJ}; ice melted =168336=0.5 kg= \dfrac{168}{336} = 0.5\text{ kg}.
  3. Ice left =0.1 kg= 0.1\text{ kg}, water =0.5 kg= 0.5\text{ kg}, temperature stays 0∘C0^{\circ}\text{C}: ice : water =1:5= 1:5.
  4. A and D are true; B, C, E are false.
  5. So the answer is (d) A and D only.
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