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NVAMediumJEE 2023Calorimetry & Change of State

JEE Physics 2023 Question with Solution

Heat energy of 184 kJ184 \, \text{kJ} is given to ice of mass 600 g600 \, \text{g} at −12∘C-12^\circ C. Specific heat of ice is 2222.3 J/kg/∘C2222.3 \, \text{J/kg/}^\circ \text{C} and latent heat of ice is 336 kJ/kg336 \, \text{kJ/kg}.

Statements: (A) Final temperature of the system will be 0∘C0^\circ C. (B) Final temperature of the system will be greater than 0∘C0^\circ C. (C) The final system will have a mixture of ice and water in the ratio of 5:15:1. (D) The final system will have a mixture of ice and water in the ratio of 1:51:5.

Find the number of correct statement(s).

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: Heat supplied is 184 kJ184 \, \text{kJ}, mass of ice is 0.600 kg0.600 \, \text{kg}, initial temperature is −12∘C-12^\circ C, specific heat of ice is 2222.3 J/kg/∘C2222.3 \, \text{J/kg/}^\circ \text{C}, and latent heat of fusion is 336000 J/kg336000 \, \text{J/kg}.

Find: The final state of the system.

First raise the temperature of ice from −12∘C-12^\circ C to 0∘C0^\circ C:

Q1=mSΔTQ_1 = mS\Delta T Q1=0.600×2222.3×12Q_1 = 0.600 \times 2222.3 \times 12 Q1=16000.56 JQ_1 = 16000.56 \, \text{J}

Remaining heat after bringing the ice to 0∘C0^\circ C is:

ΔQ1=184000−16000.56=167999.44 J\Delta Q_1 = 184000 - 16000.56 = 167999.44 \, \text{J}

Heat required to melt all the ice is:

Q2=mL=0.600×336000Q_2 = mL = 0.600 \times 336000 Q2=201600 JQ_2 = 201600 \, \text{J}

Since 167999.44 J<201600 J167999.44 \, \text{J} < 201600 \, \text{J}, all the ice does not melt.

Mass of ice melted is:

Mass of melted ice=ΔQ1L=167999.44336000≈0.4999 kg\text{Mass of melted ice} = \frac{\Delta Q_1}{L} = \frac{167999.44}{336000} \approx 0.4999 \, \text{kg}

Mass of remaining ice is:

0.600−0.4999=0.1001 kg0.600 - 0.4999 = 0.1001 \, \text{kg}

Therefore,

Ice:Water=0.1001:0.4999≈1:5\text{Ice:Water} = 0.1001 : 0.4999 \approx 1:5

Therefore, the final system is a mixture of ice and water at 0∘C0^\circ C, so the correct statements are A and D. The source the solution says A, but the working clearly supports A and D only, which corresponds to the combined answer.

Option Mapping from Statements

From the working, the final temperature remains 0∘C0^\circ C, so statement A is true and statement B is false. The ratio obtained is 1:51:5, so statement D is true, while statement C is false. Statement E is false because some ice remains. Hence the combined correct choice is A and D only.

Common mistakes

  • Using all the supplied heat directly for melting is incorrect because some heat is first needed to raise the ice temperature from −12∘C-12^\circ C to 0∘C0^\circ C. Always account for sensible heating before phase change.

  • Concluding that the final temperature becomes greater than 0∘C0^\circ C is incorrect when some ice still remains. During melting, the temperature stays at 0∘C0^\circ C until all the ice has melted.

  • Reversing the ice-to-water ratio is a common mistake. After finding melted mass and remaining ice mass, write ice : water, not water : ice.

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