Physics · Thermodynamics

JEE Main 2026 — 24 January, Morning Shift — Question 27

Density of water at 4∘C4^{\circ} \mathrm{C} and 20∘C20^{\circ} \mathrm{C} are 1000 kg/m31000 \mathrm{~kg} / \mathrm{m}^{3} and respectively. The increase in internal energy of 4 kg water when it is heated from 4∘C4{ }^{\circ} \mathrm{C} to 20∘C20^{\circ} \mathrm{C} is ____\_\_\_\_ J. (Specific heat capacity of water =4.2 J/kg=4.2 \mathrm{~J} / \mathrm{kg}. and 1 atmospheric pressure =105 Pa=10^{5} \mathrm{~Pa} )

  1. Option A:

    315826.2

  2. Option B:

    234699.2

  3. Option C:

    258700.8

  4. Option D:

    268799.2

    Correct

Answer: D

Step-by-step solution

Q=mSΔT=4×4200×16 J=268800 J\mathrm{Q}=\mathrm{mS} \Delta \mathrm{T}=4 \times 4200 \times 16 \mathrm{~J}=268800 \mathrm{~J} W=PΔV\mathrm{W}=\mathrm{P} \Delta \mathrm{V} ΔV=(mρf−mρi)=4[1998−11000]\Delta \mathrm{V}=\left(\frac{\mathrm{m}}{\rho_{\mathrm{f}}}-\frac{\mathrm{m}}{\rho_{\mathrm{i}}}\right)=4\left[\frac{1}{998}-\frac{1}{1000}\right] P=105 Pa\mathrm{P}=10^{5} \mathrm{~Pa}. ∴W=105×4×[1998−11000]=8×105103×998≈0.8 J\therefore \mathrm{W}=10^{5} \times 4 \times\left[\frac{1}{998}-\frac{1}{1000}\right]=\frac{8 \times 10^{5}}{10^{3} \times 998} \approx 0.8 \mathrm{~J} ΔU=Q−W=268799.2 J\Delta \mathrm{U}=\mathrm{Q}-\mathrm{W}=268799.2 \mathrm{~J}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy
Density of water at 4 ° C and 20 ° C are 1000 kg / m 3 and… | JEE Main 2026 PYQ with Solution · DhiX AI