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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2026 · 24 January · Morning Shift · Physics · Thermodynamics

Density of water at 4 ^ C and 20^ C are 1000 ~kg / m^ 3 and 998 ~kg...

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 998 kg/m3998 \mathrm{~kg} / \mathrm{m}^{3} respectively. The increase in internal energy of 4 kg of 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})

Options

  1. A234699.2
  2. B315826.2
  3. C258700.8
  4. D268799.2

Correct answer

D. 268799.2

Step-by-step solution

Heat absorbed by water: Q=mcΔT=4×4200×16=268800Q = mc\Delta T = 4 \times 4200 \times 16 = 268800 J.
Volume change: V1=4/1000=0.004V_1 = 4/1000 = 0.004 m³ at 4°C, V2=4/998≈0.00401V_2 = 4/998 ≈ 0.00401 m³ at 20°C.
Work done by water at constant pressure: W=PatmΔV=105×0.00001=0.8W = P_{atm}\Delta V = 10^5 \times 0.00001 = 0.8 J.
By first law of thermodynamics: ΔU=Q−W=268800−0.8=268799.2\Delta U = Q - W = 268800 - 0.8 = 268799.2 J
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