JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Heat & Thermodynamics

Density of water at

4^{\circ} C

and

2 0^{\circ} C

are

1000 kg / m^{3}

and

998 kg / m^{3}

respectively. The increase in internal energy of 4 kg water when it is heated from

4^{\circ} C

2 0^{\circ} C

is ______ J.
(Specific heat capacity of water

= 4200 J / kg^{\circ} C

and 1 atmospheric pressure

= 1 0^{5} Pa

)

Answer: (D)

Solution

By the first law of thermodynamics

Δ U = Q - W

. The heat supplied is

Q = m c Δ T = 4 \times 4200 \times 16 = 268800

J. The water expands by

Δ V = m (\frac{1}{ρ _{f}} - \frac{1}{ρ _{i}}) = 4 (\frac{1}{998} - \frac{1}{1000}) \approx 8.0 \times 1 0^{- 6} m^{3}

, so the work done against the atmosphere is

W = P Δ V = 1 0^{5} \times 8.0 \times 1 0^{- 6} \approx 0.8

J. The increase in internal energy is

268800 - 0.8 = 268799.2

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option