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MCQEasyJEE 2025Calorimetry & Change of State

JEE Physics 2025 Question with Solution

An amount of ice of mass 10−310^{-3} kg and temperature −10∘C-10^\circ C is transformed to vapor of temperature 110∘C110^\circ C by applying heat. The total amount of work required for this conversion is,

(Take, specific heat of ice = 2100 J kg−1K−12100 \, \text{J kg}^{-1} \text{K}^{-1}, specific heat of water = 4180 J kg−1K−14180 \, \text{J kg}^{-1} \text{K}^{-1}, specific heat of steam = 1920 J kg−1K−11920 \, \text{J kg}^{-1} \text{K}^{-1}, Latent heat of ice = 3.35×105 J kg−13.35 \times 10^5 \, \text{J kg}^{-1}, Latent heat of steam = 2.25×106 J kg−12.25 \times 10^6 \, \text{J kg}^{-1})

  • A

    3022 J3022 \, \text{J}

  • B

    3043 J3043 \, \text{J}

  • C

    3003 J3003 \, \text{J}

  • D

    3024 J3024 \, \text{J}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Mass of ice m=10−3 kgm = 10^{-3} \, \text{kg}, initial temperature −10∘C-10^\circ \text{C}, final state is vapour at 110∘C110^\circ \text{C}.

Find: The total heat required for the complete conversion.

The heat is added in five steps: heating ice, melting ice, heating water, vaporising water, and heating steam.

ΔQ1=mSiΔT=10−3×2100×10=21 J\Delta Q_1 = m S_i \Delta T = 10^{-3} \times 2100 \times 10 = 21 \, \text{J} ΔQ2=mLi=10−3×3.35×105=335 J\Delta Q_2 = m L_i = 10^{-3} \times 3.35 \times 10^5 = 335 \, \text{J} ΔQ3=mSwΔT=10−3×4180×100=418 J\Delta Q_3 = m S_w \Delta T = 10^{-3} \times 4180 \times 100 = 418 \, \text{J} ΔQ4=mLw=10−3×2.25×106=2250 J\Delta Q_4 = m L_w = 10^{-3} \times 2.25 \times 10^6 = 2250 \, \text{J} ΔQ5=mSsΔT=10−3×1920×10=19.2 J\Delta Q_5 = m S_s \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \, \text{J}

Therefore,

ΔQtotal=21+335+418+2250+19.2=3043.2 J\Delta Q_{\text{total}} = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \, \text{J}

Thus, the total work required is 3043 J3043 \, \text{J} approximately. The correct option is B.

Stepwise Heating Curve Approach

Given: The substance goes from ice at −10∘C-10^\circ \text{C} to vapour at 110∘C110^\circ \text{C}.

Find: Total heat for the full phase-change path.

Heating curve marked in five stages from minus ten degree Celsius to zero, zero to hundred, and hundred to one hundred ten degree Celsius.

From the heating sequence, the five energy contributions are taken one by one.

ΔQ1=m×Si×ΔT=10−3×2100×10=21 J\Delta Q_1 = m \times S_i \times \Delta T = 10^{-3} \times 2100 \times 10 = 21 \, \text{J} ΔQ2=m×Lf=10−3×3.35×105=335 J\Delta Q_2 = m \times L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \, \text{J} ΔQ3=m×Sw×ΔT=10−3×4180×100=418 J\Delta Q_3 = m \times S_w \times \Delta T = 10^{-3} \times 4180 \times 100 = 418 \, \text{J} ΔQ4=m×Lv=10−3×2.25×106=2250 J\Delta Q_4 = m \times L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \, \text{J} ΔQ5=m×Ss×ΔT=10−3×1920×10=19.2 J\Delta Q_5 = m \times S_s \times \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \, \text{J}

Adding all terms,

Q=ΔQ1+ΔQ2+ΔQ3+ΔQ4+ΔQ5Q = \Delta Q_1 + \Delta Q_2 + \Delta Q_3 + \Delta Q_4 + \Delta Q_5 Q=21+335+418+2250+19.2=3043.2 JQ = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \, \text{J}

Hence, the required heat is 3043 J3043 \, \text{J} approximately, so the correct option is B.

Common mistakes

  • Using 20∘C20^\circ \text{C} instead of 10∘C10^\circ \text{C} for heating the ice is wrong because the temperature changes from −10∘C-10^\circ \text{C} to 0∘C0^\circ \text{C} only. Use ΔT=10∘C\Delta T = 10^\circ \text{C} for the ice-heating step.

  • Using 110∘C110^\circ \text{C} instead of 10∘C10^\circ \text{C} for heating the steam is incorrect because steam is heated only from 100∘C100^\circ \text{C} to 110∘C110^\circ \text{C}. Use ΔT=10∘C\Delta T = 10^\circ \text{C} in the final step.

  • Omitting the latent heat terms gives a much smaller answer because phase changes at 0∘C0^\circ \text{C} and 100∘C100^\circ \text{C} require energy without temperature rise. Include both melting and vaporisation heats separately.

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