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MCQMediumJEE 2026Calorimetry & Change of State

JEE Physics 2026 Question with Solution

10 , \text{kg} of ice at −10∘C-10^\circ C is added to 100 , \text{kg} of water to lower its temperature from 25∘C25^\circ C. Consider no heat exchange to surroundings. The decrement in the temperature of water is _____ ∘C^\circ C. (Specific heat of ice =2100 Jkg−1 ⁣∘C−1=2100\,J kg^{-1}\!^\circ C^{-1}, specific heat of water =4200 Jkg−1 ⁣∘C−1=4200\,J kg^{-1}\!^\circ C^{-1}, latent heat of fusion of ice =3.36×105 Jkg−1=3.36\times10^5\,J kg^{-1})

  • A

    1515

  • B

    1010

  • C

    11.611.6

  • D

    6.676.67

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

  • Mass of ice mi=10 kgm_i = 10 \, \text{kg} at −10∘C-10^\circ C
  • Mass of water mw=100 kgm_w = 100 \, \text{kg} at 25∘C25^\circ C
  • Specific heat of ice sice=2100 J kg−1 ∘C−1s_{\text{ice}} = 2100 \, \text{J kg}^{-1} \, ^\circ \text{C}^{-1}
  • Specific heat of water swater=4200 J kg−1 ∘C−1s_{\text{water}} = 4200 \, \text{J kg}^{-1} \, ^\circ \text{C}^{-1}
  • Latent heat of fusion L=3.36×105 J kg−1L = 3.36 \times 10^5 \, \text{J kg}^{-1}

Find: The decrement in the temperature of water.

The ice first warms from −10∘C-10^\circ C to 0∘C0^\circ C, then melts, and finally the melted water warms from 0∘C0^\circ C to the final temperature TfT_f.

Heat gained by ice in warming to 0∘C0^\circ C:

Q1=mi⋅sice⋅ΔT=10×2100×10=2.1×105 JQ_1 = m_i \cdot s_{\text{ice}} \cdot \Delta T = 10 \times 2100 \times 10 = 2.1 \times 10^5 \, \text{J}

Heat gained in melting:

Q2=mi⋅L=10×3.36×105=3.36×106 JQ_2 = m_i \cdot L = 10 \times 3.36 \times 10^5 = 3.36 \times 10^6 \, \text{J}

Heat gained by melted ice warming to TfT_f:

Q3=mi⋅swater⋅Tf=10×4200×Tf=42000Tf JQ_3 = m_i \cdot s_{\text{water}} \cdot T_f = 10 \times 4200 \times T_f = 42000T_f \, \text{J}

So, total heat absorbed by ice is

Qabsorbed=Q1+Q2+Q3=3.57×106+42000TfQ_{\text{absorbed}} = Q_1 + Q_2 + Q_3 = 3.57 \times 10^6 + 42000T_f

Heat lost by water cooling from 25∘C25^\circ C to TfT_f:

Qlost=mw⋅swater⋅(25−Tf)=100×4200×(25−Tf)=420000(25−Tf)Q_{\text{lost}} = m_w \cdot s_{\text{water}} \cdot (25 - T_f) = 100 \times 4200 \times (25 - T_f) = 420000(25 - T_f)

Using conservation of energy,

Qabsorbed=QlostQ_{\text{absorbed}} = Q_{\text{lost}} 3.57×106+42000Tf=420000(25−Tf)3.57 \times 10^6 + 42000T_f = 420000(25 - T_f) 3570000+42000Tf=10500000−420000Tf3570000 + 42000T_f = 10500000 - 420000T_f 462000Tf=6930000462000T_f = 6930000 Tf=6930000462000=15∘CT_f = \frac{6930000}{462000} = 15^\circ \text{C}

Therefore, the decrement in the temperature of water is

ΔTwater=25−Tf=25−15=10∘C\Delta T_{\text{water}} = 25 - T_f = 25 - 15 = 10^\circ \text{C}

The correct option is B. The solution states option D, but the worked calculation gives 10∘C10^\circ \text{C}, which matches option B.

Common mistakes

  • Ignoring the heat required to melt the ice is incorrect because a large amount of energy is spent as latent heat at 0∘C0^\circ C. Always include warming of ice, melting, and then warming of the melted water.

  • Equating the water's heat loss only to warming the ice from −10∘C-10^\circ C to the final temperature is wrong because the phase change occurs in between. Break the ice process into separate stages before writing energy balance.

  • Taking the decrement directly as the final temperature is incorrect. First find TfT_f, then compute the decrease in water temperature as 25−Tf25 - T_f.

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