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Physics · Thermal Properties of Matter

JEE Main 2026 — 28 January, Morning Shift — Question 25

10 kg of ice at −10∘C-10^{\circ} \mathrm{C} is added to 100 kg of water to lower its temperature from 25∘C25^{\circ} \mathrm{C}. Consider no heat exchange to surroundings. The decrement to the temperature of water is ____\_\_\_\_ ∘C{ }^{\circ} \mathrm{C}. (specific heat of ice =2100 J/Kg.∘C=2100 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}, specific heat of water =4200 J/Kg.∘C=4200 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}, latent heat of fusion of ice =3.36×105 J/Kg=3.36 \times 10^{5} \mathrm{~J} / \mathrm{Kg} )

  1. Option A:

    10

    Correct
  2. Option B:

    15

  3. Option C:

    6.67

  4. Option D:

    11.6

Answer: A

Step-by-step solution

10×3.36×105+10×2100×10+10×4200×(T−0)=100×4200×(25−T)10 \times 3.36 \times 10^{5}+10 \times 2100 \times 10+10 \times 4200 \times(\mathrm{T}-0) =100 \times 4200 \times(25-\mathrm{T}) ⇒T=15∘C\Rightarrow \mathrm{T}=15^{\circ} \mathrm{C} ΔT=25−15=10∘C\Delta \mathrm{T}=25-15=10^{\circ} \mathrm{C}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter
Topic
Thermometry and Calorimetry
10 kg of ice at -10 ° C is added to 100 kg of water to lower its… | JEE Main 2026 PYQ with Solution · DhiX AI